2 years ago

Collin did the work to see if 10 is a solution to the equation r/4=2.5

Mathematics

1 answer:

Natasha2012 [34]2 years ago

3 0

Answer:

Yes, because if you substitute 10 for r in the equation and simplify, you find that the equation is true.

Step-by-step explanation:

$\frac{r}{4} = 2.5 \\ lhs = \frac{r}{4} \\ = \frac{10}{4} \\ = 2.5 \\ = rhs \\ \therefore \: \frac{r}{4} = 2.5 \\$

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Evaluate -5g - 6 for g = -2

Feliz [49]

Plug in g equals -5 x -2 which equals 10 because 2 negatives are a positive then 10-6 equals 4

The anwser is 4

4 0

3 years ago

Read 2 more answers

Can someone help me in this plssss I really need it

kompoz [17]

Answer:

$\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}$

Step-by-step explanation:

Your expression is

$\sqrt [3] {-54x^{5}y^{6}}$

Here's how I would simplify it.

$\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}$

$\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}$