Envision Math Accelerated Grade 7 Volume 1 Chapter 5 Generate Equivalent Expressions
We have to sort the expressions given.
An expression is a mathematical phrase, which consists of variables and constants.
We can sort the given expression into two groups.
One group would contain the expressions which have brackets and the other group will have expressions without brackets.
The first group of expressions will be
8p + 2p − 8
10p − 8
5p + 8 + 3p
10p + 4 + 2p + 2
3p + 4p + 6 + 5p
The second group of expressions will be
8 (p + 1)
2 (5p − 4)
4 (2p + 2)
3 (4p + 2)
The expression are sorted into two groups, one with expressions having brackets and the other without brackets.
We have to sort the expressions given.
An expression is a mathematical phrase, which consists of variables and constants.
We can sort the given expression into two groups.
We can also group the expression with equivalent expressions in them.
In one group, we will put the expressions that can be made one from another by multiplying or extracting a number.
The first group of expressions will be
10p − 8
2 (5p − 4) = 10
8p + 2p − 8 = 10p − 8
The second group of expressions will be
5p + 8 + 3p = 8p + 8
8 (p + 1) = 8p + 8
4 (2p + 2) = 8p + 8
The third group of expressions will be
3 (4p + 2) = 12p + 6
3p + 4p + 6 + 5p = 12p + 6
10p + 4 + 2p + 2 = 12p + 6
The expression can be grouped based on mutually equivalent expressions.
Question. How can algebraic expressions be used to represent and solve problems.
We have to tell that how can algebraic expressions be used to represent and solve problems.
The combination of variables and constants, in mathematics, is referred to as an expression.
The expression having algebraic terms are called as algebraic expressions.
These expressions are helpful in expressing the problems which consists of variables.
The variables are substituted with values given and then evaluated to obtain the solution or desired result.
The algebraic expressions represent the problems which can be evaluated by substituting values in the expression.
Question. How much it would cost to rent a scooter for 3 1/2 hours and watercraft for 1 ¾ hours.
We have to determine how much it would cost to rent a scooter for 3 1/2 hours and watercraft for 1 ¾ hours.
When we solve or evaluate an expression it means that we substitute and replace some values in the place of variables, to get the required result.
The order of operation that will be followed is Brackets, multiplication or division, and addition or subtraction.
The expressions are
s = \(3 \frac{1}{2}\)
= 3.5
w=\(1 \frac{3}{4}\)
= 1.75
We will evaluate as below:
15.5s + 22.8w = x
15.5cot 3.5 + 22.8.1.75 = x
54.25 + 39.9 = x
x = 94.15
The renting of both these amounts of time will cost $94.15.
Question. Misumi used to determine her account balance after “w” week. She gets 3.5 dollars in a day and 8.74 per hour.
We have to give an expression for Misumi used to determine her account balance after “w” week.
She gets 3.5 dollars in a day and 8.74 per hour.
We consider “h” as the number of hours.
h = \(15\frac{1}{2}\)
The order of operation that will be followed is.
Brackets, multiplication or division, addition or subtraction.
We evaluate the expression as below:
3.5 + 8.74⋅h = x
3.5 + 8.74.5.5 = x
3.5 + 48.07 = x
x = 51.57
She will receive the amount of $51.57 in that day.
Question. How should we determine which value to use for the constant and which value to use for the coefficient.
We have to tell that how should we determine which value to use for the constant and which value to use for the coefficient.
An expression is a mathematical phrase, which consists of variables and constants.
We determine the amount after w weeks as the below expression
217 + 25.5.w
In the given expression, w is the coefficient as the number of weeks can be changed.
And the constant amount is 25.5 which she always deposits.
Thus, in a problem, the factor whose value changes in different situations is considered as a variable, as its value varies.
While the value which remains the same is taken as constant.
The values which vary are considered as variables and the ones which are the same are determined as constants.
Question. How can algebraic expressions be used to represent and solve problems.
We have to tell that how can algebraic expressions be used to represent and solve problems.
The combination of variables and constants, in mathematics, is referred to as an expression.
The expression having algebraic terms are called as algebraic expressions.
These expressions are helpful in expressing the problems which consist of variables.
The variables are substituted with values given and then evaluated to obtain the solution or desired result.
The algebraic expressions represent the problems which can be evaluated by substituting values in the expression.
Question. Explain how is a constant term different than a variable term for an expression that represents a real-world situation.
We need to explain how is a constant term different than a variable term for an expression that represents a real-world situation.
The constant term is nothing but a term that is constant i.e., the value doesn’t change no matter what.
The variable term is the one which is varying based on different values.
As the name suggests, it is a variable one.
For example, in the expression, 5x + 2 = 0
The term 2 is a constant since it doesn’t change no matter what.
The term x is a variable it changes based on the input.
The value of the constant doesn’t change while that of the variable changes based on its input.
Question. Explain why we can have different values when evaluating an algebraic expression.
We need to explain why we can have different values when evaluating an algebraic expression.
When evaluating an algebraic expression, we may have some different values.
But the end result is always the same. This is because we can choose which operation we can do first.
For example, both addition and subtraction are commutative.
If we do addition first, the values will be different.
Or if we do subtraction first, the values will be different.
But the end result obtained by the expression will be the same.
We can choose which operation we can choose first. This is why we can have different values when evaluating an algebraic expression.
Question. A tank containing 35 gallons of water is leaking at a rate of \(\frac{1}{4}\) gallon per minute. Write an expression to determine the number of gallons left in the tank after m minutes.
Given that, A tank containing 35 gallons of water is leaking at a rate of \(\frac{1}{4}\) gallon per minute.
We need to write an expression to determine the number of gallons left in the tank after m minutes.
Here, the rate is a variable since it varies every minute.
The amount of water is constant.
We need to find the number of gallons left after m minutes.
Thus, the expression becomes
35−\(\frac{1}{4}\) m
The number of gallons left in the tank after m minutes will be 35−\(\frac{1}{4}\) m
Question. Write an algebraic expression that Marshall can use to determine the total cost of buying a watermelon that weighs w pounds and some tomatoes that weigh t pounds. Find how much will it cost to buy a watermelon that weighs \(18\frac{1}{2}\) pounds and 5 pounds of tomatoes.
We need to write an algebraic expression that Marshall can use to determine the total cost of buying a watermelon that weighs w pounds and some tomatoes that weigh t pounds.
We need to find how much will it cost to buy a watermelon that weighs \(18 \frac{1}{2}\) pounds and 5 pounds of tomatoes.
The total of buying will be represented by the expression
0.68w + 3.25t
Given that, t = 5, w = \(18 \frac{1}{2}\)
Substituting this we get
0.68w + 3.25t = 0.68 × \(18 \frac{1}{2}\) + 3.25 × 5
= 0.68 × \( \frac{37}{2}\) + 16.25
= 12.58 + 16.25
= 28.83
He have to pay $28.83 altogether.
Question. Find the value of \(\frac{3}{8}\)x-4.5 when the value of x = 0.4.
The value of \( \frac{3}{8}\)x−4.5 when the value of x = 0.4
Substituting x = 0.4 in the given expression, we get
\( \frac{3}{8}\)x − 4.5
= \( \frac{3}{8}\) × 0.4−4.5
= \( \frac{3}{2}\) × 0.1−4.5
= 0.15 − 4.5
= −4.35
The value of \( \frac{3}{8}\)x − 4.5 = −4.35
Question. Find the value of 8.4n-3.2p when n = 2 and p = 4.
The value of 8.4n−3.2p when n = 2 and p = 4
Substituting the value of n = 2, p = 4 , we get
8.4n − 3.2p = 8.4(2)−3.2(4)
= 16.8 − 12.8
= 4
The value of 8.4n−3.2p = 4
Question. Write an expression that represents the height of a tree that began at 6 feet and increases by 2 feet per year.
We need to write an expression that represents the height of a tree that began at 6 feet and increases by 2 feet per year.
Let y represent the number of years.
Here, the height of the tree initially is 6 feet.
The increase in height will be 2 feet per year.
Here, y represents the number of years.
Thus, the expression will be
6 + 2y
The expression that represents the height of a tree will be 6 + 2y
Question. Evaluate the expression for the value 3d-4 of the variable (s).
We need to evaluate the given expression for the given value of the variable(s).
The given expression is 3d−4
The value of d = 1.2
Substituting the value of the variable in the expression, we get
3d−4 = 3(1.2) − 4
= 3.6 − 4
= −0.4
The value of 3d−4=−0.4
Question. Evaluate the expression for the value 0.5f – 2.3g of the variable(s).
We need to evaluate the given expression for the given value of the variable(s).
The given expression is 0.5f−2.3g
The value of f = 12, g = 2
Substituting the value of the variable in the expression, we get
0.5f − 2.3g = 0.5(12) − 2.3(2)
= 6−4.6
= 1.4
The value of 0.5f − 2.3g = 1.4
Question. Evaluate the given expression for the value \(\frac{2}{3}\) p+3 of the variable (s).
We need to evaluate the given expression for the given value of the variable(s).
The given expression is \( \frac{2}{3}\) p+3
The value p= \( \frac{3}{5}\)
Substituting the value of the variable in the expression, we get
\(\frac{2}{3} p+3=\frac{2}{3}\left(\frac{3}{5}\right)+3\)= \(\frac{2}{5}+3\)
= \(\frac{2+15}{5}\)
= \(\frac{17}{5}\)
The value of \( \frac{2}{3}\) p + 3 = \( \frac{17}{5}\) p + 3
Question. Evaluate the expression for the value 34 + \(\frac{4}{9}\)w of the variable(s).
We need to evaluate the given expression for the given value of the variable(s).
The given expression is 34 + \( \frac{4}{9}\)w
The value of w = − \( \frac{1}{2}\)
Substituting the value of the variable in the expression, we get
\(34+\frac{4}{9} w=34+\frac{4}{9}\left(\frac{-1}{2}\right)\)= \(34-\frac{2}{9}\)
= \(\frac{306-2}{9}\)
= \(\frac{304}{9}\)
= 33.78
The value of 34 + \( \frac{4}{9}\)w = 33.78
Question. Find the expression that can be used to determine the total cost of buying g pounds of granola for $3.25 per pound and f pounds of flour for $0.74 per pound.
We need to find the expression that can be used to determine the total cost of buying g pounds of granola for $3.25 per pound and f pounds of flour for $0.74 per pound.
The cost of buying granola is 3.25 × g
The cost of buying flour is 0.74 × f
Therefore, the total cost of buying both will be
3.25 × g + 0.74 × f = 3.25g + 0.74f
The expression that can be used to determine the total cost will be 3.25g + 0.74f
Question. Find the expression that can be used to determine the total weight of a box that by itself weighs 0.2 kilogram and contains p plaques that weigh 1.3 kilograms each.
We need to find the expression that can be used to determine the total weight of a box that by itself weighs 0.2 kilogram and contains p plaques that weigh 1.3 kilograms each.
The weight of the box will be 0.2kg
The number of plaques will be p
The weight of each plaques will be 1.3kg
Thus, the expression will be
Total weight of the box = 0.2 +1.3p
The expression that can be used to determine the total weight of a box is (A) 1.3p + 0.2
Question. The expression -120 + 13m represents a submarine that began at a depth of 120 feet below sea level and ascended at a rate of 13 feet per minute. Find the depth of the submarine after 6 minutes.
Given that, the expression −120 + 13m represents a submarine that began at a depth of 120 feet below sea level and ascended at a rate of 13 feet per minute.
We need to find the depth of the submarine after 6 minutes.
Here, the given expression is −120 + 13m
Given that, m = 6
Substituting it, we get
−120 + 13m =−120 + 13(6)
= −120 + 78
= −42
The depth of the submarine after 6 minutes will be −42
Question. The capacity is 3000 ft3 The expression to determine the amount of grain left.
The capacity is 3000 ft3
The rate is \(\frac{3.5 f t^3}{s}\)
The expression to determine the amount of grain left will be
3000 − 3.5s
Here, s is the number of seconds.
The expression to determine the amount of grain left will be 3000 − 3.5s
Question. The expression 5 – 5x to have a negative value, we need to find what must be true about the value of x.
For the expression 5− 5x to have a negative value, we need to find what must be true about the value of x.
The given expression is 5− 5x
The value will be negative when the value of 5x is greater than the number 5.
If x is negative, thus x < 0 the expression becomes positive.
If x = 0 then the expression becomes positive.
If x = 1 then the expression becomes 5−5(1) = 5−5 = 0
If x > 1 then the expression becomes negative.
x >1 must be true about the value of x.
Question. The outside temperature was 73°F at 1 P.M and decreases at a rate of 1.5°F each hour. Find the expression that can be used to determine the temperature h hours after 1 P.M.
Given that, the outside temperature was 73 °F at 1 P.M and decreases at a rate of 1.5 °F each hour.
We need to find the expression that can be used to determine the temperature h hours after 1 P.M.
The rate is a variable that varies every hour.
The outside temperature initially was constant.
The temperature decreases thus we need to subtract the temperature from the outside temperature.
The expression that can be used to determine the temperature h hours after 1 P.M will be, 73 − 1.5h