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3 years ago

11

Help help , Please help! Brainliest if correct! What are the end behaviors of f(x)=-2(x-17)^4

1 answer:

lions [1.4K]3 years ago

5 0

Answer:

D

Step-by-step explanation:

When this function is graphed, both ends go down forever

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Twin brothers, person A and person B, can mow their grandparent's lawn together in 40 minutes. Person A could mow the lawn by

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22 minutes I believe

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3 years ago

Which quadratic equation defines the function that has zeros at − 1/12 and 1/4 ?

Cerrena [4.2K]

$\bf \begin{cases}
x=-\frac{1}{2}\implies &x+\frac{1}{2}=0\\\\
x=\frac{1}{4}\implies &x-\frac{1}{4}=0
\end{cases}
\\\\\\
\left( x+\frac{1}{2} \right)\left( x-\frac{1}{4} \right)=\stackrel{y}{0}\implies \stackrel{FOIL}{x^2+\frac{1}{4}x-\frac{1}{8}}=y$

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4 years ago

Please help me worth a bunch of points 26 points

Romashka [77]

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3 years ago

Read 2 more answers

Judy simplified the expression (3x + 1)(x – 3) – (2 – 2x2 – 3x). What is the constant term in the simplified expression?

andrey2020 [161]

Answer:

Step-by-step explanation:

(3x+1)(x-3)-(2-2x^2-3x)

3x^2+x-9x-3-2+2x^2+3x

3x^2+2x^2+x-9x+3x-3-2

5x^2-5x-5

the constant term in the simplified expression is -5.

5 0

3 years ago

In triangle ΔABC, ∠C is a right angle and CD is the height to

Zina [86]

Answer:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

Step-by-step explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles $m\angle CDB=m\angle CDA=90$ °

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

$m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha$

From ΔCBD,

$m\angle CBD$ is same as $m\angle B$ .

So, $m\angle CBD=90-\alpha$

$m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha$

Now, from ΔCAD,

$m\angle CAD$ is same as $m\angle A$

So, $m\angle CAD=\alpha$

$m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha$

Hence, the unknown angles of both the triangles are:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

5 0

3 years ago