Ask question

Ask question

All categories

3 years ago

11

Graph the function and find the range

1 answer:

kramer3 years ago

6 0

The answer for the graph would be C. While x is less than or equal to 0, there is the graph of -4x. While x is greater than x, there is the graph of -5 (horizontal line). Furthermore, the closed dot at the end of the -4x represents the less than or EQUAL TO. The dot at the end of the -5 is open because it is only true when x > 0.

The range (y-values where the graph exists) is [-5]∪[0, infinity)

You might be interested in

Use a half-angle identity to find the exact value

Tatiana [17]

Given:

$\cos 15^{\circ}$

To find:

The exact value of cos 15°.

Solution:

$$\cos 15^{\circ}=\cos\frac{ 30^{\circ}}{2}$

Using half-angle identity:

$$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}$

$$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}$

Using the trigonometric identity: $\cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}$

$$=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}$

Let us first solve the fraction in the numerator.

$$=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}$

Using fraction rule: $\frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}$

$$=\sqrt{\frac {2+\sqrt{3}}{4}}$

Apply radical rule: $\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}$

$$=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}$

Using $\sqrt{4} =2$ :

$$=\frac{\sqrt{2+\sqrt{3}}}{2}$

$$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}$

5 0

3 years ago

PLEASE HELP WILL MARK BRAINLIEST!

pav-90 [236]

Answer:

1, 5, 17, 53, 161

Step-by-step explanation:

Given:

$a_1=1$

$a_n=3a_{n-1}+2$

First five terms of the sequence:

$a_1=1$

$a_2=3a_1+2=3(1)+2=5$

$a_3=3a_2+2=3(5)+2=17$

$a_4=3a_3+2=3(17)+2=53$

$a_5=3a_4+2=3(53)+2=161$

7 0

2 years ago

Read 2 more answers

Solve and i give brainliest

Temka [501]

Answer:

0.25

Step-by-step explanation:

-9/6 divided by 3/-2 = 0.25

i am sorry

4 0

2 years ago

Read 2 more answers

How was abcdefgh transform to a'b'c'd'e'g'h'

strojnjashka [21]

It reflects across the x axis

8 0

3 years ago

Read 2 more answers

For the funtion f(x)= (x+7)^5, Find f^-1 (x)

quester [9]

Answer:

The inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

Step-by-step explanation:

For the function $f(x)= (x+7)^5$ , Find $f^{-1} (x)$

For finding inverse of x,

First let:

$y=(x+7)^5$

Now replace x with y and y with x

$x=(y+7)^5$

Now, solve for y

Taking 5th square root on both sides

$\sqrt[5]{x}=\sqrt[5]{(y+7)^5}\\\sqrt[5]{x}=y+7\\=> y+7=\sqrt[5]{x}\\y=\sqrt[5]{x}-7$

Now, replace y with $f^{-1} (x)$

$f^{-1} (x)=\sqrt[5]{x}+7$

So, the inverse of function $f(x)= (x+7)^5$ is $\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}$

Option A is correct option.

6 0

3 years ago