A rectangular prism has a volume of 90 cm^3 choose all the possible sets of lengths widths and heights for the rectangular prism

Please help!!! ASAP the bottom number is 3 btw.

Nataliya [291]

Answer:

5*3=15

Step-by-step explanation:

3 0

3 years ago

|a+3| ÷(a−3) , if a=−17 Please HELP

spin [16.1K]

Answer:

-0.7

Explanation:

All you do is substitute x.

Add/Subtract from left to right.

|-17 + 3| / (-17 - 3)

Find the absolute value of |-14|.

|-14| / (-20)

Divide.

14 / (-20)

-0.7

5 0

3 years ago

Given r(x) = 11/ (x - 42)

julsineya [31]

For a given function f(x) we define the domain restrictions as values of x that we can not use in our function. Also, for a function f(x) we define the inverse g(x) as a function such that:

g(f(x)) = x = f(g(x))

The restriction is:

x ≠ 4

The inverse is:

$y = 4 + \sqrt{\frac{11}{x} }$

Here our function is:

$f(x) = \frac{11}{(x - 4)^2}$

We know that we can not divide by zero, so the only restriction in this function will be the one that makes the denominator equal to zero.

(x - 4)^2 = 0

x - 4 = 0

x = 4

So the only value of x that we need to remove from the domain is x = 4.

To find the inverse we try with the general form:

$g(x) = a + \sqrt{\frac{b}{x} }$

Evaluating this in our function we get:

$g(f(x)) = a + \sqrt{\frac{b}{f(x)} } = a + \sqrt{\frac{b*(x - 4)^2}{11 }}\\\\g(f(x)) = a + \sqrt{\frac{b}{11 }}*(x - 4)$

Remember that the thing above must be equal to x, so we get:

$g(f(x)) = a + \sqrt{\frac{b}{11 }}*(x - 4) = x\\\\{\frac{b}{11 }} = 1\\{\frac{b}{11 }}*4 - a = 0$

From the two above equations we find:

b = 11

a = 4

Thus the inverse equation is:

$y = 4 + \sqrt{\frac{11}{x} }$

If you want to learn more, you can read:

brainly.com/question/10300045

3 0

2 years ago

PLEASE HELP ANSWER THIS QUESTION I DON'T KNOW IT AT ALL

Lena [83]

<h2>PLEASE MARK BRAINLIEST!</h2>

Answer:

Is y = 4x - 7 a linear function?
Is y = 6x² - 1 a linear function?
Is y = $\frac{1}{2x}$ + 10 a linear function?

Step-by-step explanation:

Yes it is - when you graph this equation, it results in a [straight] line, signalling that it is a linear function.
No it's not - when you graph this equation, it results in a v- kinda shape on the graph. Linear functions are [straight] lines on a graph, and this line wasn't straight. In fact, this wasn't even a line.
No it's not - when you graph this equation, it results in a bend at the origin. The line on the graph is not straight, so this is not a linear equation.

For the graphs -