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3 years ago

Welp here a pick bc I really have no idea what Im looking at

Mathematics

1 answer:

mylen [45]3 years ago

7 0

Answer:

D) $-\sqrt[3]{x-1}$

Step-by-step explanation:

-the function is the cube root

-the cube root of zero is zero

- the graph is at 0 when x = 1, thus "x-1" must be part of answer

- the cube root of -8 is -2, at -8 the graph is +2. so the function is negated

$-\sqrt[3]{x-1}$

You might be interested in

What is the height of a rectangle with an area of 105 and a base of 15

BaLLatris [955]

Answer:

Step-by-step explanation:

Please let me know if you want me to add an explanation as to why this is the answer. I can definitely do that, I just don’t want to waste my time in case you don’t want me to :)

8 0

3 years ago

Evaluate the difference quotient for the given function. Simplify your answer.

nasty-shy [4]

I suppose you mean

$f(x) = \dfrac{x+5}{x+1}$

Then

$f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2$

and the difference quotient is

$\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}$

If it's the case that <em>x</em> ≠ 3, then (<em>x</em> - 3)/(<em>x</em> - 3) reduces to 1, and you would be left with

$\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}$

4 0

2 years ago

In parallelogram DEFG, DH equals X +3, HF equals 3Y, GH equals 2X -5 and HE equals 5Y plus to find the values of X and Y

daser333 [38]

The values of X and Y are 30 and 11 respectively

<h3>How to determine the values of X and Y?</h3>

The figure that represents the complete question is added as an attachment

The given parameters are:

DH = X +3

HF = 3Y

GH = 2X -5

HE = 5Y

From the attached parallelogram, we have:

DH = HF

GH = HE

Substitute the known values in the above equation

X + 3 = 3Y

2X - 5 = 5Y

Make X the subject in X + 3 = 3Y

X = 3Y - 3

Substitute X = 3Y - 3 in 2X - 5 = 5Y

2(3Y - 3) - 5 = 5Y

Expand

6Y - 6 - 5 = 5Y

Evaluate the like terms

Y = 11

Substitute Y = 11 in X = 3Y - 3

X = 3*11 - 3

Evaluate

X = 30

Hence, the values of X and Y are 30 and 11 respectively