3 years ago

What are the domain and range of each relation?

Mathematics

2 answers:

almond37 [142]3 years ago

4 0

Answer:

it's the second one

Step-by-step explanation:

the domain is your x and the range are your y int

Alenkasestr [34]3 years ago

4 0

The third (last) box

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I WILL GIVE GOLD MEDALS PLEASE HELP ??? Can somebody help with these 2 math questions?

taurus [48]

4)

$\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta\pi r^2}{360}\qquad 
\begin{cases}
\theta=\textit{angle in degrees}\\
r=radius\\
----------\\
r=CE=9\\
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\end{cases}$

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3 years ago

Describe how to transform the quantity of the fifth root of x to the seventh power, to the third power into an expression with a

MrRa [10]

Fifth root of x is x^1/5 as an exponential expression. . This to the 7th power is

x ^(1/5 * 7) which is x^(7/5). Bringing this to the third power is x^ (21/5) which is your answer.

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3 years ago

What is the simplified expression for the expression below?

Viktor [21]

Answer:

-2x + 3

Step-by-step explanation:

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3 years ago

Mike just hopped on the edge of a merry-go-round. What are his linear and angular speeds if the diameter of the merry-go-round i

dsp73

Step-by-step explanation:

Given that,

The diameter of the merry-go-round, d = 14 feet

Time taken, t = 6 seconds

Radius, r = 7 feet

The linear speed of the merry-go-round is given by :

$v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 7}{6}\\\\=7.33\ m/s$

Also,

$v=r\omega$

Where

$\omega$ is the angular speed

So,

$\omega=\dfrac{v}{r}\\\\\omega=\dfrac{7.33}{7}\\\\=1.04\ rad/s$

Hence, his linear and angular speeds are 7.33 m/s and 1.04 rad/s.

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3 years ago

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Arturiano [62]

B. 39

Good luck on yourself timed exam!

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3 years ago