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Evaluate Algebraic Expressions – ProblemsEvaluate Algebraic Expressions – Problems
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Question 1 of 5
1. Question
Find the velocity `(v)` using the formula `v=u+at`, given the following values:`u=45``a=2``t=3`- `\text(v)` = (51)
Hint
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Fantastic!
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Order of Operations (PEMDAS)
P`=`Parentheses
E`=`Exponents
M/D`=`Multiplication/Division
A/S`=`Addition/SubtractionSubstitute the values to the given velocity formula, then solve.`u` `=` `45` `a` `=` `2` `t` `=` `3` `v` `=` `u``+``a``t` `=` `45``+``(2)``(3)` `=` `45+6` `=` `51` `51` -
Question 2 of 5
2. Question
Find the value of `M` using the formula `M=(10N)/(P+5)`, given the following values:`N=9``P=4`- `\text(M)` = (10)
Hint
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Excellent!
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Order of Operations (PEMDAS)
P`=`Parentheses
E`=`Exponents
M/D`=`Multiplication/Division
A/S`=`Addition/SubtractionSubstitute the values to the given formula for `M`, then solve.`N` `=` `9` `P` `=` `4` `M` `=` $$\frac{10\color{#9a00c7}{N}}{\color{#00880A}{P}+5}$$ `=` $$\frac{10(\color{#9a00c7}{9})}{\color{#00880A}{4}+5}$$ `=` `90/9` `=` `10` `10` -
Question 3 of 5
3. Question
Find the area of the rectangle below given that: `A=LxxW`
- (40) `\text(cm)^2`
Hint
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Label the values in the rectangle.`L` (length) `=` `10 \text(cm)` longer side `W` (width) `=` `4 \text(cm)` shorter side Substitute the values to the given formula, then solve.Area `=` `L``xx``W` `=` `10``xx``4` `=` `40 \text(cm)^2` `cm` is squared since we multiplied the values `40 \text(cm)^2` -
Question 4 of 5
4. Question
Find the perimeter of the rectangle below given that: `P=2(L+W)`
- (38) `\text(cm)`
Hint
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Correct!
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Label the values in the rectangle.`L` (length) `=` `12 \text(cm)` longer side `W` (width) `=` `7 \text(cm)` shorter side Substitute the values to the given formula, then solve.`P` `=` `2(``L``+``W``)` `=` `2(``12``+``7``)` `=` `2(19)` `=` `38 \text(cm)` `38 \text(cm)` -
Question 5 of 5
5. Question
Find the area of the rectangle below given that: `m=5cm`
- (80) `\text(cm)^2`
Hint
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Area of a Rectangle
`\text(Area )=``\text(width)``times``\text(length)`Label the values in the rectangle.width `=` `m` shorter side length `=` `2m+6` longer side Substitute the values to the area formula, then solve given that `m=5`.`\text(Area)` `=` width`times`length `=` `m``times``(2m+6)` `=` `5``xx(2xx``5` `+6)` Substitute `m=5` `=` `5xx(10+6)` `=` `5xx16` `=` `80 \text(cm)^2` `80 \text(cm)^2`
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