{"id":25030,"date":"2018-06-01T10:09:50","date_gmt":"2018-06-01T10:09:50","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=25030"},"modified":"2018-06-01T10:09:50","modified_gmt":"2018-06-01T10:09:50","slug":"algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/","title":{"rendered":"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1"},"content":{"rendered":"

Algebra 1 Common Core Answers\u00a0Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1<\/strong><\/span><\/h2>\n

Algebra 1 Common Core Answers Student Edition Grade 8 – 9<\/a><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 1LC<\/strong>
\n(a) <\/strong>Given is to describe an expression as algebraic or numerical:-
\nTo describe above expression as algebraic or numerical, we first define both algebraic and numerical expressions as below:
\n7\u00f72
\nAlgebraic expression-<\/strong> It is a mathematical phrase which includes one or more variables.
\nFor example<\/strong>:- 6n-8, 3h, \\( \\frac { x }{ 3 } \\)
\nNumerical expression-<\/strong> It is a mathematical phrase which includes numbers and operations symbols, but no variables.
\nFor example<\/strong>:- 7+2, 3\u00d76-3, \\( \\frac { 14 }{ 7 } \\)
\nClearly, the above expression 7\u00f72 is a numerical expression as it includes no variable and just includes two numbers (7 and 2 )and one operational sign that is division (\u00f7) and the value of this expression always remain constant.
\nThus 7\u00f72 is a numerical expression.
\n(b) <\/strong>4m+6
\nTo describe above expression as algebraic or numerical, we first define both algebraic and numerical expressions which is stated below:
\nAlgebraic expression-<\/strong> It is a mathematical phrase which includes one or more variables.
\nFor example<\/strong>:- 6n-8,3h,\\( \\frac { x }{ 3 } \\)
\nNumerical expression-<\/strong> It is a mathematical phrase which includes numbers and operations symbols, but no variables
\nFor example<\/strong>:- 7+2, 3\u00d76-3, \\( \\frac { 14 }{ 7 } \\)
\nClearly, the above expression is an algebraic expression as it includes one variable (m) and the value of this expression keeps on changing with the change of value of variable.
\nThus 4m+6 is a algebraic expression.
\n(c) <\/strong>2(5-4)
\nTo describe above expression as algebraic or numerical, we first define both algebraic and numerical expressions which is stated below:
\nAlgebraic expression-<\/strong> It is a mathematical phrase which includes one or more variables.
\nFor example<\/strong>:- 6n-8,3h,\\( \\frac { x }{ 3 } \\)
\nNumerical expression-<\/strong> It is a mathematical phrase which includes numbers and operations symbols, but no variables
\nFor example<\/strong>:- 7+2, 3\u00d76-3, \\( \\frac { 14 }{ 7 } \\)
\nClearly, the above expression is a numerical expression 2(5-4) as it includes no variable and just includes two numbers (2,5 and 4) and one operational sign that is division (\u00d7 and -) and the value of this expression always remain constant.
\nThus 2(5-4) is a algebraic expression.<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 2LC<\/strong>
\na) <\/strong>Given is to write an algebraic expression for the following phrase:-
\nThe product of 9 and a number t =9t
\nIt means to multiply 9 with t
\nHence, the product of 9 and a number t = 9t
\nb) <\/strong>Given is to write an algebraic expression for the following phrase:-
\nThe difference of a number x and \\( \\frac { 1 }{ 2 } \\)
\nIt means to subtract \\( \\frac { 1 }{ 2 } \\) from x
\nHence, the difference of a number x and \\( \\frac { 1 }{ 2 } \\) \\( =\\quad x-\\frac { 1 }{ 2 } \\)
\nc) <\/strong>Given is to write an algebraic expression for the following phrase:-
\nThe sum of a number m and 7.1
\nIt means to add 7.1 to m
\nHence, the sum of a number m and 7.1 = m+7.1
\nd) <\/strong>Given is to write an algebraic expression for the following phrase:-
\nThe quotient of 207 and a number n
\nIt means to divide 207 by n
\nHence, the quotient of 207 and a number n = \\( \\frac { 207 }{ n } \\)<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 3LC<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n6c<\/strong>
\nClearly, 6c means the product of 6 and a number c
\nThus, the phrase for 6c is the product of 6 and a number c<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 4LC<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\nx-1<\/strong>
\nClearly, x-1 is the difference of a number x and 1
\nThus, the phrase for x-1 is the difference of a number x and 1<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 5LC<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n\\( \\frac { t }{ 2 } \\)<\/strong>
\nClearly, \\( \\frac { t }{ 2 } \\) is the quotient of a number t and 2
\nThus, the phrase for \\( \\frac { t }{ 2 } \\) is the quotient of a number t and 2<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 6LC<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n3t – 4<\/strong>
\nFirstly, it means the product of 3 and a number t
\nThen, clearly it means the 4 less than a product of 3 and t
\nThus, the phrase for 3t-4 is 4 less than a product of 3 and t<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 7LC<\/strong>
\nGiven to write a difference between numerical expression and algebraic expression:
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-7lc\"<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 8LC<\/strong>
\nGiven is to decide which expression out of given two expressions represent total cost to rent a truck by considering the given below table:
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-8lc\"
\nThe given expressions are:-
\n49n+0.75 or 49+0.75n<\/strong>
\nAfter considering the value of the cost from the above table for different number of miles, it is understood that,
\n$49 is constant for any number of miles.<\/strong>
\n$75 kepps on changing with number of miles.<\/strong>
\nFor n number of miles, the expression will become:
\n$49+($.75\u00d7n)
\n=$49+($.75\u00d7n)
\n=$49+$.75\u00d7n
\nThe above expression for n number of miles can also be written as:
\n=49+0.75n
\nThus, 49+0.75n is the expression that represents total cost to rent a truck that he drives n miles.<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 9E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\n4 more than p
\nIt means to add 4 to p
\nHence, 4 more than p = 4+p<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 10E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\ny minus 12
\nIt means to subtract 12 from y
\nHence, y minus 12 = y-12<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 11E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\nThe quotient of n and 8
\nIt means to divide n by 8
\nHence, the quotient of n and 8 = \\( \\frac { t }{ 2 } \\) or n\u00f78<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 12E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\nThe product of 15 and c
\nIt means to multiply 15 with c
\nHence, The product of 15 and c = 15\u00d7c<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 13E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\nA number t divided by 82
\nIt means to divide t by 82
\nHence, a number t divided by 82 = \\( \\frac { t }{ 82 } \\) or t\u00f782<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 14E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\nThe sum of 13 and twice a number h
\nFirstly, it means multiply 2 with h which is equal to 2\u00d7h or 2h
\nThen, add 13 to this result
\nHence, 13 more than twice a number h = 13+2h<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 15E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\n6.7 more than the product of 5 and n
\nFirstly, it means multiply 5 with n, which is equal to 5\u00d7n or 5n
\nThen, add 6.7 to this result
\nHence, 6.7 more than the product of 5 and n = 6.7 + 5n<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 16E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\n9.85 less than the product of 37 and t
\nFirstly, it means multiply 37 with t which is equal to 37\u00d7t or 37t
\nThen, subtract 9.85 from this result
\nHence, 9.85 less than the product of 37 and t = 37t – 9.85<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 17E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\nq+5<\/strong>
\nClearly, it means the sum of a number q and 5
\nThus, the phrase for q+5<\/strong> is the sum of a number q and 5<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 18E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n\\( \\frac { y }{ 5 } \\)<\/strong>
\nClearly, it means the quotient of a number y and 5
\nThus, the phrase for \\( \\frac { y }{ 5 } \\) is the quotient of a number y and 5<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 19E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n12x<\/strong>
\nClearly, it means the product of 12 and a number x
\nThus, the phrase for\u00a012x<\/strong> is the product of 12 and a number x<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 20E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n49+m<\/strong>
\nClearly, it means the sum of 49 and a number m
\nThus, the phrase for 49+m<\/strong> is the sum of 49 and a number m<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 21E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n9n+1<\/strong>
\nFirstly, it means the product of 9 and a number n
\nThen, clearly it means the sum of 1 and 9 times a number n
\nThus, the phrase for 9n+1<\/strong> is the sum of 1 and 9 times a number n<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 22E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n\\( \\frac { z }{ 8 } -9 \\)<\/strong>
\nFirstly, it means the quotient of a number z and 8
\nThen, clearly it means 9 less than quotient of a number z and 8
\nThus, the phrase for \\( \\frac { z }{ 8 } -9 \\) is 9 less than quotient of a number z and 8<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 23E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n\\( 15-\\frac { 1\\cdot 5 }{ d } \\)<\/strong>
\nFirstly, it means the quotient of 1.5 and a number d
\nThen, clearly it means the difference of 15 and the quotient of 1.5 and a number d
\nSo, the phrase for \\( 15-\\frac { 1\\cdot 5 }{ d } \\) is the difference of 15 and the quotient of 1.5 and a number d<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 24E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\n2(5-n)<\/strong>
\nFirstly, it means the difference of 5 and a number n
\nThen, clearly it means the product of 2 and difference of 5 and a number n
\nThus, the phrase for 2(5-n)<\/strong> is the product of 2 and difference of 5 and a number n<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 25E<\/strong>
\nGiven is to write a rule in words and as an algebraic expression to model the relationship as per given below table:-
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-25E\"
\nAs per given question, he buys a bicycle and helmet on rent.
\nHe pays $9<\/strong> for each hour he uses it and $5<\/strong> for helmet.
\nAs per above given table,
\nWhen he uses bicycle for 1 hour, the rental cost will become<\/strong>
\n(Rental cost of bicycle \u00d7 number of hours) + rental cost helmet<\/strong>
\nwhere number of hour = 1<\/strong>
\nSimilarly, when he uses bicycle for 2 hours, the rental cost will become
\n(Rental cost of bicycle \u00d7 number of hours) + rental cost helmet<\/strong>
\nwhere number of hour = 2<\/strong>
\nNext, when he uses bicycle for 3 hours, the rental cost will become
\n(Rental cost of bicycle \u00d7 number of hours) + rental cost helmet<\/strong>
\nwhere number of hour = 3<\/strong>
\nFinally, when he uses bicycle for n hours, the rental cost will become
\n(Rental cost of bicycle \u00d7 number of hours) + rental cost helmet<\/strong>
\nwhere number of hour = n<\/strong>
\nSo, the expression for rental cost of bicycle for n number of hours = ($9\u00d7n)+$5<\/strong>
\nThe expression ($9\u00d7n)+$5<\/strong> in words can be stated as follows:-
\nFirstly, it means the product of 9 and a number n.
\nThen, add 5 to this result.
\nThus, the phrase for ($9\u00d7n)+$5<\/strong> is 5 more than the product of a number n and 9<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 26E<\/strong>
\nGiven is to write a rule in words and as an algebraic expression to model the relationship as per given below table:-
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-26E\"
\nAs per given question, a salesperson earned a weekly salary of $150<\/strong> and also paid $2<\/strong> for each pair of shoes he or she sold during the week.
\nAs per above given table,
\nWhen he or she sold 5 pair of shoes, the total earned will become
\nweekly salary + (extra amount he or she paid for selling number of pair of shoes \u00d7 number of pair of shoes)<\/strong>
\nwhere number of pair of shoes sold = 5<\/strong>
\nSimilarly, when he or she sold 10 pair of shoes, the total earned will become
\nweekly salary + (extra amount he or she paid for selling number of pair of shoes \u00d7 number of pair of shoes)<\/strong>
\nwhere number of pair of shoes sold = 10<\/strong>
\nNext, when he or she sold 15 pair of shoes, the total earned will become
\nweekly salary + (extra amount he or she paid for selling number of pair of shoes \u00d7 number of pair of shoes)<\/strong>
\nwhere number of pair of shoes sold = 15<\/strong>
\nFinally, when he or she sold n pair of shoes, the total earned will become
\nweekly salary + (extra amount he or she paid for selling number of pair of shoes \u00d7 number of pair of shoes)<\/strong>
\nwhere number of pair of shoes sold = n<\/strong>
\nSo, the expression for total earned when he or she sold n number of pair of shoes
\n=$150+($2\u00d7n)<\/strong>
\nThe expression =$150+($2\u00d7n)<\/strong> in words can be stated as follows:-
\nFirstly, it means the product of 2 and a number n.
\nThen, add 150 to this result.
\nThus, the phrase for $150+($2\u00d7n)<\/strong> is 150 more than the product of 2 and a number n<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 27E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\n8 minus the product of 9 and r
\nFirstly, it means multiply 9 with r which is equal to 9\u00d7r or 9r
\nThen, subtract 8 from this result
\nHence, 8 minus the product of 9 and r = 9r-8<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 28E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\nThe sum of 15 and x, plus 7
\nFirstly, it means add 15 and a number x which is equal to 15+x
\nThen, add 7 to this result
\nHence, the sum of 15 and x, plus 7 = (15+x)+7<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 29E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\n4 less than three seventh of y
\nFirstly, it means multiply a number y with \\( \\frac { 3 }{ 7 } \\) which is equal to \\( \\frac { 3 }{ 7 } \\times y \\) or \\( \\frac { 3 }{ 7 } y \\)
\nThen, subtract 4 from this result
\nHence, 4 less than three seventh of \\( y=\\frac { 3 }{ 7 } y-4 \\)<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 30E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\nThe quotient of 12 and the product of 5 and t
\nFirstly, it means multiply 5 and a number t which is equal to 5\u00d7t or 5t
\nThen, divide 12 by this result
\nHence, the quotient of12 and the product of 5 and \\( t=\\frac { 12 }{ 5t } \\)<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 31E<\/strong>
\nGiven expression is:-
\n\\( \\frac { 5 }{ n} \\)
\nFor this expression, word phrase written by a student is:-
\nThe quotient of n and 5
\nClearly, the phrase written by him was incorrect as his phrase would represent the division of n and 5 which is given by:-
\n\\( \\frac { n }{ 5} \\)
\nSo, it is incorrect as per given expression.
\nNow, the given expression, that is, \\( \\frac { 5 }{ n} \\) represent the division of 5 and n
\nSo, the correct phrase for the given expression is the quotient of 5 and n<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 32E<\/strong>
\nGiven is to write an algebraic expression to give the rule for following statement:-
\nThe number of bagels in any number b of baker\u2019s dozen
\nBelow is the given table showing number of bagels a shop gave him per baker\u2019s dozen:
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-32E\"
\nAs per above given table,
\nWhen baker\u2019s dozen is 1, the number of bagels will become
\n13\u00d7number of bangels,where number of bangels = 1
\nSimilarly, when baker\u2019s dozen is 2, the number of bagels will become
\n13\u00d7number of bangels,where number of bangels = 2
\nNext, when baker\u2019s dozen is 3, the number of bagels will become
\n13\u00d7number of bangels,where number of bangels = 3
\nFinally, when baker\u2019s dozen is n, the number of bagels will become
\n13\u00d7number of bangels,where number of bangels = b
\nSo, the algebraic expression for number of bagels in b number of baker\u2019s dozen = 13b
\nThe pattern of increase in the number of bagels is described as:-
\nAs per given table, the number of bagels increased with the multiple of 13 as
\nFor 1 baker\u2019s dozen, the numbers of bagels are 13 which is 13\u00d71
\nFor 2 baker\u2019s dozen, the numbers of bagels are 26 which is 13\u00d72
\nFor 3 baker\u2019s dozen, the numbers of bagels are 13 which is 13\u00d73
\nAnd so on up to for b baker\u2019s dozen, the numbers of bagels are 13b which is 13\u00d7b
\nThus, pattern of increase in the number of bagels is multiple of 13<\/strong>
\nOperation on b to find the number of bagels is given by:-
\nConsider the expression to determine the operation on b to calculate the number of bagels 13b
\nClearly, the above expression means multiply 13 with the number b to find number of bagels
\nThus, multiplication operation on b is performed to find number of bagels<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 33E<\/strong>
\nEach ticket cost is $4.50<\/strong>.
\na.<\/strong> If n is the number of tickets purchased, Write an expression that gives the total cost of buying n<\/strong> tickets.
\nThe expression is 4.50m<\/strong>.
\nb.<\/strong> Suppose the total cost for n tickets is $36<\/strong>.Find the total cost if one more ticket is purchased.
\nThe total cost is $36 + $4.50 = $40.5<\/strong>.<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 34E<\/strong>
\nGiven is to write an algebraic expression for the following statement:-
\nNumber of boxes he will have wrapped when she has wrapped x number of boxes.
\nShe and he wrapped gift boxes at the same pace.
\nIn the first figure, only she wrapped 2 gift boxes.
\nIn the second figure, she has wrapped 3 gift boxes and he has wrapped 1 gift box.
\nIn the third figure, she has wrapped 4 gift boxes and he has wrapped 2 gift boxes.
\nAs per above given statements, the expression will become
\nnumber of boxes that he was wrapped = number of boxes that she has wrapped – 2<\/strong>
\nNow, when she has wrapped x number of boxes, thus it means he has wrapped x – 2
\nHence, the expression for number of boxes he wrapped when she wrapped x boxes = x – 2<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 35E<\/strong>
\nGiven is to write an expression that gives the value in dollars of d dimes:-
\nThat is, we want to find the following conversion
\nd dimes = ? dollars<\/strong>
\nNow, as \\(1dime=\\frac { 1 }{ 10 } dollars\\)
\nSo, for d dimes multiply d on both sides, then expression will become,
\n\\( d\\quad dimes=d\\times \\frac { 1 }{ 10 } dollars \\)
\nd dimes = d \u00d7 0.10 dollars
\nOn multiplying, it becomes
\nd dimes = 0.10d dollars
\nThus, the expression that gives the value in dollars of d dimes = 0.10d
\nHence, option A is correct<\/strong>.<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 36E<\/strong>
\nGiven is to describe a real world situation and representation of variable for the following expression:
\n5t<\/strong>
\nThe above expression 5t means the product of 5 and t.
\nIf we take the variable t as the cost of 1kg of apples, then by above definition, 5t represents the cost of 5kg of apples.
\nThus, in a real world situation 5t represents cost of 5kg of apples and t represents the cost of 1kg of apples<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 37E<\/strong>
\nGiven is to describe a real world situation and representation of variable for the following expression:
\nb+3
\nThe above expression b+3 means the 3 more than b.
\nIf the variable b represents as the age of a person X, then by above definition, b+3 represents the age of person Y who is 3 years elder to X.
\nThus, in a real world situation b+3 represents age of person Y who is 3 years elder to person X and b represents the age of person X<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 38E<\/strong>
\nGiven is to describe a real world situation and representation of variable for the following expression:
\n\\( \\frac { 40 }{ h} \\)
\nThe above expression \\( \\frac { 40 }{ h} \\) means the quotient of 40 and a number h.
\nIf the variable h represents the number of persons, then by above definition, \\( \\frac { 40 }{ h} \\) represents the amount of money received by 1 person when a total money of $40 is distributed among h number of persons.
\nThus, in a real world situation \\( \\frac { 40 }{ h} \\) represents the amount of money received by 1 person when a total money of is distributed $40 among h number of persons and h represents number of persons<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 39E<\/strong>
\nConsider the following two expressions:
\n(5-2) \u00f7 n and (5\u00f7n) – 2 ……(1)
\n(5\u00f7n) – 2 ……(2)
\nConsider the following phrase:
\n2 less than 5 divided by a number n
\nFirstly, it means to divide 5 by a number n that is 5\u00f7n and then, subtract 2 from the result.
\nClearly, expression for the given phrase is as follows:
\n(5\u00f7n) – 2
\nSo, the expression (2) is a suitable expression for the given phrase.
\nThe given expression (5-2) \u00f7 n describes as follows:
\nIt means to subtract 2 from 5 and then, divide the result by n.
\nThe expression (1) is not a suitable expression for the given phrase.
\nTherefore, it can be concluded that,
\nThe phrase is correctly represented by the expression (5\u00f7n) – 2 and not by (5-2) \u00f7 n<\/strong>
\nNext, is to decide whether verbal description lack precision.
\nYes, verbal description lack precision as they do not provide stress on the order of operations which is needed to be performed in a specific way to get the correct answer.<\/strong>
\nIn the above phrase 2 less than 5 divided by a number n, it is not clear whether one has to apply subtraction first and then division or vice-versa.<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 40E<\/strong>
\nGiven is to expression that could represent the given diagram:-
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-40E\"
\nThe above diagram is the tabular representation of a relationship between number x and 1
\nThe diagram can be represented in two ways by counting the total number of 1\u2019s either column wise (in a vertical direction) or row wise (in a horizontal direction)
\nNumber of columns in the given figure =4
\nNumber of rows in the given figure =3
\nFirst, we count the number of 1\u2019s column wise.
\nClearly in each column there are three 1\u2019s.
\nSo, counting 1\u2019s in first column =3
\nAs, each column has same number of 1\u2019s and there are 4 columns
\nThus, total number of 1\u2019s in the given diagram =3+3+3+3
\nAdding we get,
\nTotal number of 1\u2019s in the given diagram =12
\nSecondly, we count the number of 1\u2019s row wise.
\nClearly in each row there are four 1\u2019s.
\nSo, counting 1\u2019s in first row =4
\nAs, each row has same number of 1\u2019s and there are 3 rows.
\nThus, total number of 1\u2019s in the given diagram =4+4+4
\nAdding we get,
\nTotal number of 1\u2019s in the given diagram =12
\nThus, the two different expressions to represent the given diagram are as follows:
\nFirstly, 3+3+3+3 and secondly 4+4+4<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 41E<\/strong>
\nGiven is to expression that could represent the given diagram:-
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-41E\"
\nThe above diagram is the tabular representation of a relationship between number x and 1
\nThe diagram can be represented in two ways by counting the total number of 1\u2019s either column wise (in a vertical direction) or row wise (in a horizontal direction)
\nNumber of columns in the first row =4
\nNumber of columns in the second row =2
\nFirst, we count the number of 1\u2019s column wise.
\nIn first column, there are two 1\u2019s
\nIn second column, there are two 1\u2019s
\nIn third column, there is only one 1
\nIn fourth column, there is only one 1
\nSo, counting 1\u2019s in first column =2
\nCounting 1\u2019s in second column =2
\nCounting 1\u2019s in third column =1
\nCounting 1\u2019s in fourth column =1
\nThus, total number of 1\u2019s in the given diagram =2+2+1+1
\nAdding we get,
\nTotal number of 1\u2019s in the given diagram =6
\nSecondly, we count the number of 1\u2019s row wise.
\nIn first row, there are 4 number of 1\u2019s.
\nIn second row, there are 2 number of 1\u2019s.
\nThus, total number of 1\u2019s in the given diagram =4+2
\nAdding we get,
\nTotal number of 1\u2019s in the given diagram =6
\nThus, the two different expressions to represent the given diagram are as follows:
\nFirstly, 2+2+1+1 and secondly 4+2<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 42E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\n2 less than the product of 3 and a number x
\nFirstly, it means to multiply 3 with x which is equal to 3\u00d7x or 3x
\nThen, subtract 2 from this result
\nThus, 2 less than the product of 3 and a number x = 3x-2
\nHence, option A is correct<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 43E<\/strong>
\nGiven is to write a phrase for the following algebraic expression:-
\nn\u00f78
\nWe can also write above expression as \\( \\frac { n }{ 8} \\)
\nClearly, it means the quotient of a number n and 8
\nThus, phrase for n\u00f78 is the quotient of a number n and 8
\nHence, option G is correct<\/strong>.<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 44E<\/strong>
\nGiven is to write an algebraic expression for the following phrase:-
\nA state park charges are entrance fees + $18 for each night of camping
\nwhere an entrance fee is $20
\nBelow is the given table showing the relationship for an entrance fees and charges for each night:-
\n\"algebra-1-common-core-answers-chapter-1-1-foundations-for-algebra-exercise-1-1-44E\"
\nBy considering the above table, the expression for total cost of n nights of camping is:-
\n$18\u00d7n+$20
\n=$18n+$20
\nOr we can write an expression as 18n+20
\nHence, option B is correct<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 45E<\/strong>
\nGiven expression is:
\n\\( \\frac { 1 }{ 4 } +\\frac { 1 }{ 2 } \\) \u2026\u2026 (1)
\nFirst, we find LCD of the denominators of all above fractions
\nListing all denominators: 4 and 2
\nPrime factors of 4
\n=2\u00d72
\nPrime factors of 2 = 2\u00d71
\nSo, LCD of 4 and 2 =2\u00d72
\nThus, LCD of \\( \\frac { 1 }{ 4 } \\) , \\( \\frac { 1 }{ 2 } \\) is 4
\nNext, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is, 4.
\n\\( \\frac { 1 }{ 4 } \\) is already in the equivalent form.
\nNext, to convert \\( \\frac { 1 }{ 2 } \\)
\nput \\( \\frac { 1 }{ 2 } =\\frac { ? }{ 4 } \\)
\nTo find the missing numerator, we observe the relation between denominators that is 2 and 4.
\nClearly, \\( \\frac { 1 }{ 2 } =\\frac { 1\\times 2 }{ 2\\times 2 } \\)
\n\\( =\\frac { 2 }{ 4 } \\)
\nNow writing the equivalent form of both the fractions in above expression (1), it becomes
\n\\( \\frac { 1 }{ 4 } +\\frac { 1 }{ 2 } =\\frac { 1 }{ 4 } +\\frac { 2 }{ 4 } \\)
\nNow, both the above fractions have same denominator that is 4. So they all are like fractions and to combine them, we just combine their numerators, keeping denominator same
\n\\( \\frac { 1 }{ 4 } +\\frac { 1 }{ 2 } =\\frac { 1+2 }{ 4 } \\)
\n\\( \\frac { 1 }{ 4 } +\\frac { 1 }{ 2 } =\\frac { 3 }{ 4 } \\)
\nMaking prime factors of numerator and denominator, it becomes
\n\\( \\frac { 3\\times 1 }{ 2\\times 2 } \\)
\nCancelling the same factors in numerator and denominator, as it has no same factors then it becomes
\n\\( =\\frac { 3 }{ 4 } \\)
\nHence,
\n\\( \\frac { 1 }{ 4 } +\\frac { 1 }{ 2 } =\\frac { 3 }{ 4 } \\)<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 46E<\/strong>
\nGiven expression is:
\n\\( \\frac { 9 }{ 14 } -\\frac { 2 }{ 7 } \\) \u2026\u2026 (1)
\nFirst, we find LCD of the denominators of both the above fractions
\nPrime factors of 14 = 2\u00d77
\nPrime factors of 7 = 7\u00d71
\nSo, LCD of \\( \\frac { 9 }{ 14 } \\) and \\( \\frac { 2 }{ 7 } \\) = 2\u00d77 =14
\nNext, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is 14.
\n\\( \\frac { 9 }{ 14 } \\) is already in the equivalent form.
\nNext, to convert \\( \\frac { 2 }{ 7 } \\), put \\( \\frac { 2 }{ 7 } =\\frac { ? }{ 14 } \\)
\nClearly, \\( \\frac { 2 }{ 7 } =\\frac { 2\\times 2 }{ 7\\times 2 } \\)
\n\\( =\\frac { 4 }{ 14 } \\)
\nNow writing the equivalent form of both the fractions in above expression (1), it becomes
\n\\( \\frac { 9 }{ 14 } -\\frac { 2 }{ 7 } =\\frac { 9 }{ 14 } -\\frac { 4 }{ 14 } \\)
\nNow, both the above fractions have same denominator that is 14. So they both are like fractions and to combine them, we just combine their numerators, keeping the denominator same
\n\\( \\frac { 9 }{ 14 } -\\frac { 2 }{ 7 } =\\frac { 9-4 }{ 14 } \\)
\nOn subtraction the numerator, it becomes
\n\\( =\\frac { 5 }{ 14 } \\)
\nHence,
\n\\( \\frac { 9 }{ 14 } -\\frac { 2 }{ 7 } =\\frac { 5 }{ 14 } \\)<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 47E<\/strong>
\nGiven expression is:
\n\\( \\frac { 2 }{ 5 } +\\frac { 3 }{ 10 } \\) \u2026\u2026 (1)
\nFirst, we find LCD of the denominators of all above fractions
\nListing all denominators: 5 and 10
\nPrime factors of 5 = 5\u00d71
\nPrime factors of 10 = 2\u00d75
\nSo, LCD of 5 and 10 = 2\u00d75 = 10
\nThus, LCD of \\( \\frac { 2 }{ 5 } \\), \\( \\frac { 3 }{ 10 } \\) is 10
\nNext, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is, 10.
\nTo convert \\( \\frac { 2 }{ 5 } \\), put \\( \\frac { 2 }{ 5 } =\\frac { ? }{ 10 } \\)
\nTo find the missing numerator, we observe the relation between denominators that is 5 and 10.
\nClearly, \\( \\frac { 2 }{ 5 } =\\frac { 2\\times 2 }{ 5\\times 2 } \\)
\nNext, \\( \\frac { 3 }{ 10 } \\) is already in the equivalent form.
\nNow writing the equivalent form of two fractions in above expression (1), it becomes
\n\\( \\frac { 2 }{ 5 } +\\frac { 3 }{ 10 } =\\frac { 4 }{ 10 } +\\frac { 3 }{ 10 } \\)
\nNow, both the above fractions have same denominator that is 10. So they all are like fractions and to combine them, we just combine their numerators, keeping denominator same
\n\\( \\frac { 2 }{ 5 } +\\frac { 3 }{ 10 } =\\frac { 4+3 }{ 10 } \\)
\nOn adding the terms of numerator, it becomes
\n\\( =\\frac { 7 }{ 10 } \\)
\nMaking prime factors of numerator and denominator, it becomes
\n\\( =\\frac { 7\\times 1 }{ 2\\times 5 } \\)
\nCancelling the same factors in numerator and denominator, as there is no same factor then, it becomes
\n\\( =\\frac { 7 }{ 10 } \\)
\nHence,
\n\\( \\frac { 2 }{ 5 } +\\frac { 3 }{ 10 } =\\frac { 7 }{ 10 } \\)<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 48E<\/strong>
\nGiven expression is:
\n\\( \\frac { 5 }{ 6 } -\\frac { 2 }{ 3 } \\) \u2026\u2026 (1)
\nFirst, we find LCD of the denominators of both the above fractions
\nPrime factors of 6 = 3\u00d72
\nPrime factors of 3 = 3\u00d71
\nSo, LCD of \\( \\frac { 5 }{ 6 } \\) and \\( \\frac { 2 }{ 3 } \\) = 2\u00d73\u00d71 = 6
\nNext, we convert each of these fraction to an equivalent fraction with the denominator same as their LCD, that is 6.
\n\\( \\frac { 5 }{ 6 } \\) is already in equivalent form.
\nNext, to convert \\( \\frac { 2 }{ 3 } \\), put \\( \\frac { 2 }{ 3 } =\\frac { ? }{ 6 } \\)
\nClearly, \\( \\frac { 2 }{ 3 } =\\frac { 2\\times 2 }{ 3\\times 2 } \\)
\n\\( =\\frac { 4 }{ 6 } \\)
\nNow writing the equivalent form of both the fractions in above expression (1), it becomes
\n\\( \\frac { 5 }{ 6 } -\\frac { 2 }{ 3 } =\\frac { 5 }{ 6 } -\\frac { 4 }{ 6 } \\)
\nNow, both the above fractions have same denominator that is 6. So they both are like fractions and to combine them, we just combine their numerators, keeping denominator same
\n\\( =\\frac { 5-4 }{ 6 } \\)
\n\\( =\\frac { 1}{ 6 } \\)
\nHence,
\n\\( \\frac { 5 }{ 6 } -\\frac { 2 }{ 3 } =\\frac { 1 }{ 6 } \\)<\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 49E<\/strong>
\nGiven numbers are 3 and 6
\nPrime factors of 3 = 1\u00d73
\nPrime factors of 6 = 2\u00d73
\nBy definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 3 just occurs once in both the whole number but 2 does not occurs once in 3
\nThen, greatest common factor of 3 and 6 =3
\nHence, greatest common factor of 3 and 6 is 3<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 50E<\/strong>
\nGiven numbers are 12 and 15
\nPrime factors of 12 = 2\u00d72\u00d73
\nPrime factors of 15 = 3\u00d75
\nBy definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 3 just occur once in both the whole number but 2 and 5 do not occur once in them.
\nThen, greatest common factor of 12 and 15 = 3
\nHence, greatest common factor of 12 and 15 is 3<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 51E<\/strong>
\nGiven numbers are 7 and 11
\nPrime factors of 7 = 1\u00d77
\nPrime factors of 11 = 1\u00d711
\nBy definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 1 just occurs once in both the whole number but 7 and 11 does not occur once in them.
\nThen, greatest common factor of 7 and 11 = 1
\nHence, greatest common factor of 7 and 11 is 1<\/strong><\/p>\n

Chapter 1 Foundations for Algebra Exercise 1.1 52E<\/strong>
\nGiven numbers are 12 and 8
\nPrime factors of 12 = 2\u00d72\u00d73
\nPrime factors of 8 = 2\u00d72\u00d72
\nBy definition of greatest common factor of two whole numbers, it contains each prime factor occurring that number of times equals to the minimum number of times it appears in the each of the whole numbers. Clearly, 2 occur twice in both the whole numbers but 3 do not occur once in them.
\nThen, greatest common factor of 12 and 8 = 2\u00d72 = 4
\nHence, greatest common factor of 12 and 8 is 4<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Algebra 1 Common Core Answers\u00a0Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1 Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1 1LC (a) Given is to describe an expression as algebraic or numerical:- To describe above expression as algebraic or numerical, … Read more<\/a><\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[20593],"tags":[20595,20597,20602,20594,20599,20600,20601,20596,20598],"yoast_head":"\nAlgebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1 - CBSE Library<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1\" \/>\n<meta property=\"og:description\" content=\"Algebra 1 Common Core Answers\u00a0Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1 Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1 1LC (a) Given is to describe an expression as algebraic or numerical:- To describe above expression as algebraic or numerical, ... Read more\" \/>\n<meta property=\"og:url\" content=\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/\" \/>\n<meta property=\"og:site_name\" content=\"CBSE Library\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/aplustopper\/\" \/>\n<meta property=\"article:published_time\" content=\"2018-06-01T10:09:50+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png\" \/>\n<meta name=\"twitter:card\" content=\"summary\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Raju\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"26 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Organization\",\"@id\":\"https:\/\/cbselibrary.com\/#organization\",\"name\":\"Aplus Topper\",\"url\":\"https:\/\/cbselibrary.com\/\",\"sameAs\":[\"https:\/\/www.facebook.com\/aplustopper\/\"],\"logo\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/#logo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"contentUrl\":\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg\",\"width\":1585,\"height\":375,\"caption\":\"Aplus Topper\"},\"image\":{\"@id\":\"https:\/\/cbselibrary.com\/#logo\"}},{\"@type\":\"WebSite\",\"@id\":\"https:\/\/cbselibrary.com\/#website\",\"url\":\"https:\/\/cbselibrary.com\/\",\"name\":\"CBSE Library\",\"description\":\"Improve your Grades\",\"publisher\":{\"@id\":\"https:\/\/cbselibrary.com\/#organization\"},\"potentialAction\":[{\"@type\":\"SearchAction\",\"target\":{\"@type\":\"EntryPoint\",\"urlTemplate\":\"https:\/\/cbselibrary.com\/?s={search_term_string}\"},\"query-input\":\"required name=search_term_string\"}],\"inLanguage\":\"en-US\"},{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#primaryimage\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png\",\"contentUrl\":\"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png\"},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#webpage\",\"url\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/\",\"name\":\"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1 - CBSE Library\",\"isPartOf\":{\"@id\":\"https:\/\/cbselibrary.com\/#website\"},\"primaryImageOfPage\":{\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#primaryimage\"},\"datePublished\":\"2018-06-01T10:09:50+00:00\",\"dateModified\":\"2018-06-01T10:09:50+00:00\",\"breadcrumb\":{\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#breadcrumb\"},\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"ReadAction\",\"target\":[\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/\"]}]},{\"@type\":\"BreadcrumbList\",\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#breadcrumb\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https:\/\/cbselibrary.com\/\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1\"}]},{\"@type\":\"Article\",\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#webpage\"},\"author\":{\"@id\":\"https:\/\/cbselibrary.com\/#\/schema\/person\/bd70dda77c8f886a86b7f2142f75d9c2\"},\"headline\":\"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1\",\"datePublished\":\"2018-06-01T10:09:50+00:00\",\"dateModified\":\"2018-06-01T10:09:50+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#webpage\"},\"wordCount\":5177,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/cbselibrary.com\/#organization\"},\"image\":{\"@id\":\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png\",\"keywords\":[\"Algebra 1 Common Core answers and solutions\",\"Algebra 1 Common Core by Charles Hall and Basia Kennedy\",\"algebra 1 common core edition pdf\",\"Algebra 1 Common Core Student Edition Grade 8 - 9\",\"algebra 1 common core textbook solutions pdf\",\"algebra 1 student edition\",\"algebra 1 workbook common core standards edition answer key online\",\"ISBN: 9780133185485\",\"practice and problem solving workbook algebra 1 common core answers\"],\"articleSection\":[\"Algebra\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#respond\"]}]},{\"@type\":\"Person\",\"@id\":\"https:\/\/cbselibrary.com\/#\/schema\/person\/bd70dda77c8f886a86b7f2142f75d9c2\",\"name\":\"Raju\",\"image\":{\"@type\":\"ImageObject\",\"@id\":\"https:\/\/cbselibrary.com\/#personlogo\",\"inLanguage\":\"en-US\",\"url\":\"https:\/\/secure.gravatar.com\/avatar\/38ccf4a5413b0ede9fc79802993d6981?s=96&d=mm&r=g\",\"contentUrl\":\"https:\/\/secure.gravatar.com\/avatar\/38ccf4a5413b0ede9fc79802993d6981?s=96&d=mm&r=g\",\"caption\":\"Raju\"},\"url\":\"https:\/\/cbselibrary.com\/author\/raju\/\"}]}<\/script>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1 - CBSE Library","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/","og_locale":"en_US","og_type":"article","og_title":"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1","og_description":"Algebra 1 Common Core Answers\u00a0Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1 Algebra 1 Common Core Answers Student Edition Grade 8 – 9 Chapter 1 Foundations for Algebra Exercise 1.1 1LC (a) Given is to describe an expression as algebraic or numerical:- To describe above expression as algebraic or numerical, ... Read more","og_url":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/","og_site_name":"CBSE Library","article_publisher":"https:\/\/www.facebook.com\/aplustopper\/","article_published_time":"2018-06-01T10:09:50+00:00","og_image":[{"url":"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png"}],"twitter_card":"summary","twitter_misc":{"Written by":"Raju","Est. reading time":"26 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Organization","@id":"https:\/\/cbselibrary.com\/#organization","name":"Aplus Topper","url":"https:\/\/cbselibrary.com\/","sameAs":["https:\/\/www.facebook.com\/aplustopper\/"],"logo":{"@type":"ImageObject","@id":"https:\/\/cbselibrary.com\/#logo","inLanguage":"en-US","url":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","contentUrl":"https:\/\/cbselibrary.com\/wp-content\/uploads\/2018\/12\/Aplus_380x90-logo.jpg","width":1585,"height":375,"caption":"Aplus Topper"},"image":{"@id":"https:\/\/cbselibrary.com\/#logo"}},{"@type":"WebSite","@id":"https:\/\/cbselibrary.com\/#website","url":"https:\/\/cbselibrary.com\/","name":"CBSE Library","description":"Improve your Grades","publisher":{"@id":"https:\/\/cbselibrary.com\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/cbselibrary.com\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"ImageObject","@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#primaryimage","inLanguage":"en-US","url":"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png","contentUrl":"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png"},{"@type":"WebPage","@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#webpage","url":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/","name":"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1 - CBSE Library","isPartOf":{"@id":"https:\/\/cbselibrary.com\/#website"},"primaryImageOfPage":{"@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#primaryimage"},"datePublished":"2018-06-01T10:09:50+00:00","dateModified":"2018-06-01T10:09:50+00:00","breadcrumb":{"@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/cbselibrary.com\/"},{"@type":"ListItem","position":2,"name":"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1"}]},{"@type":"Article","@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#article","isPartOf":{"@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#webpage"},"author":{"@id":"https:\/\/cbselibrary.com\/#\/schema\/person\/bd70dda77c8f886a86b7f2142f75d9c2"},"headline":"Algebra 1 Common Core Answers\u00a0Chapter 1 Foundations for Algebra Exercise 1.1","datePublished":"2018-06-01T10:09:50+00:00","dateModified":"2018-06-01T10:09:50+00:00","mainEntityOfPage":{"@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#webpage"},"wordCount":5177,"commentCount":0,"publisher":{"@id":"https:\/\/cbselibrary.com\/#organization"},"image":{"@id":"https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#primaryimage"},"thumbnailUrl":"https:\/\/farm2.staticflickr.com\/1739\/28615874208_e6f3076a78_o.png","keywords":["Algebra 1 Common Core answers and solutions","Algebra 1 Common Core by Charles Hall and Basia Kennedy","algebra 1 common core edition pdf","Algebra 1 Common Core Student Edition Grade 8 - 9","algebra 1 common core textbook solutions pdf","algebra 1 student edition","algebra 1 workbook common core standards edition answer key online","ISBN: 9780133185485","practice and problem solving workbook algebra 1 common core answers"],"articleSection":["Algebra"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/cbselibrary.com\/algebra-1-common-core-answers-chapter-1-foundations-for-algebra-exercise-1-1\/#respond"]}]},{"@type":"Person","@id":"https:\/\/cbselibrary.com\/#\/schema\/person\/bd70dda77c8f886a86b7f2142f75d9c2","name":"Raju","image":{"@type":"ImageObject","@id":"https:\/\/cbselibrary.com\/#personlogo","inLanguage":"en-US","url":"https:\/\/secure.gravatar.com\/avatar\/38ccf4a5413b0ede9fc79802993d6981?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/38ccf4a5413b0ede9fc79802993d6981?s=96&d=mm&r=g","caption":"Raju"},"url":"https:\/\/cbselibrary.com\/author\/raju\/"}]}},"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/posts\/25030"}],"collection":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/comments?post=25030"}],"version-history":[{"count":0,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/posts\/25030\/revisions"}],"wp:attachment":[{"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/media?parent=25030"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/categories?post=25030"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/cbselibrary.com\/wp-json\/wp\/v2\/tags?post=25030"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}