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3 years ago

11

HELP ME PLSSSSS IM BEGGING UUUU!!!!!

1 answer:

Mrac [35]3 years ago

6 0

Answer:

D

Step-by-step explanation:

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Find the value of each variable in the parallelogram.

Lorico [155]

Answer:

w = 2, z = 7

Step-by-step explanation:

If you have any questions about the way I solved it, don't hesitate to ask ;)

3 0

4 years ago

HELLLLLLLLLLLLLLLLLPPPPPPPPP

ahrayia [7]

Answer:

C

Step-by-step explanation:

-6(-2/3 + 2x)

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3 0

3 years ago

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What is the primary difference between exponential and linear functions?

Yanka [14]

Answer:

A

Step-by-step explanation:

We know that the growth of a linear function will always be constant. So, that eliminates B and C.

A quadratic function can be a function such as $x^2$ , or $5x^2+7x$ , etc.

An exponential function wouldn't be $x^2$ , it would be $2^x$ ! Or $3^x$ , or $(1.738283)^x$ , etc. Therefore, D is eliminated.

So, the answer is $\boxed{A}$ and we're done!

4 0

2 years ago

What is the total area of the shape

solong [7]

Answer:

Area : 704 square units

Step-by-step explanation:

First you do 16 x 10 = 160

then you do 32 x 17 = 544

then you add 160 + 544 = 704 (D)

7 0

3 years ago

G(x) = -2x^3 – 15x^2 + 36x

shusha [124]

Consider the function $G(x) = -2x^3 - 15x^2 + 36x.$ First, factor it:

$G(x) = -2x^3 - 15x^2 + 36x=-x(2x^2+15x-36)=\\ \\=-x\cdot 2\cdot \left(x-\dfrac{-15-\sqrt{513}}{4}\right)\cdot \left(x-\dfrac{-15+\sqrt{513}}{4}\right).$

The x-intercepts are at points $\left(\dfrac{-15-\sqrt{513} }{4},0\right),\ (0,0),\ \left(\dfrac{-15+\sqrt{513} }{4},0\right).$

1. From the attached graph you can see that

function is positive for $x\in \left(-\infrty, \dfrac{-15-\sqrt{513} }{4}\right)\cup \left(0,\dfrac{-15+\sqrt{513} }{4}\right);$
function is negative for $x\in \left(\dfrac{-15-\sqrt{513} }{4},0\right)\cup \left(\dfrac{-15+\sqrt{513} }{4},\infty\right).$

2. Since

$G(-x) = -2(-x)^3 - 15(-x)^2 + 36(-x)=2x^3-15x^2-36x\neq G(x)\ \text{and }\neq -G(x)$ the function is neither even nor odd.

3. The domain is $x\in (-\infty,\infty),$ the range is $y\in (-\infty,\infty).$

8 0

3 years ago

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