Could somebody please help me??

Mathematics

2 answers:

solniwko [45]2 years ago

7 0

Answer:

She didn't make a mistake

Step-by-step explanation:

aleksandrvk [35]2 years ago

5 0

Answer:

She did not make a mistake tell me if i am right pls hope I helped

Step-by-step explanation:

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Sasha sells T-shirts. Each day she earns a set amount, plus a commission.

WARRIOR [948]

Answer:

Domain → 0 < x < 5

Step-by-step explanation:

Sasha sells T-shirts and earns a fixed amount plus a commission by selling each shirt. (As given in the table)

Table attached shows a linear function (A regular increase in total pay with the increase in number of shirts sold)

So the input values of the table (Number of shirts sold) will represent the domain of the linear function.

Hence, reasonable domain for the function will be → 0 < x < 5

5 0

3 years ago

HELP PLEASE!! I DONT UNDERSTAND!!!!!!!!!! THANKS SO MUCH

8090 [49]

Hello, please consider the following.

We will multiply the numerator and denominator by

$4+\sqrt{6x}$

to get rid of the root in the denominator.

First of all, we cannot divide by 0, right? So, we need to make sure that the denominator is different from 0.

$4-\sqrt{6x} =0\sqrt{6x}=4\\\\\text{Take the square}\\\\6x=4^2=16\\\\x=\dfrac{16}{6}=\dfrac{8}{3}$

We need to take any x real number different from 8/3 then and simplify the expression.

Let's do it!

$\begin{aligned}\dfrac{4}{4-\sqrt{6x}}&=\dfrac{4(4+\sqrt{6x})}{(4+\sqrt{6x})(4-\sqrt{6x})}\\\\&=\dfrac{4(4+\sqrt{6x})}{(4^2-\sqrt{6x}^2)}\\\\&=\dfrac{4(4+\sqrt{6x})}{(16-6x)}\\\\&=\dfrac{2(4+\sqrt{6x})}{(8-3x)}\\\\&\large \boxed{=\dfrac{8+2\sqrt{6x}}{8-3x}}\end{aligned}$