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ad-work [718]
3 years ago
9

If f(x)=x+7 and g(x)=1/x-13, what is the domain of (f0g)(x)?

Mathematics
1 answer:
aev [14]3 years ago
6 0

Answer:

Step-by-step explanation:

Hope this helps :)

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The radius of a circle is 5 miles. What is the length of a 135° arc?
SCORPION-xisa [38]

Answer:

75

Step-by-step explanation:

an arc is half a circle

7 0
2 years ago
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“7a - 17 = 4a +1”<br><br> Step by step pls
Monica [59]

Answer:

a = 6

Step-by-step explanation:

7a-17 = 4a +1 ; Subtract 4a from both sides

3a - 17 = 1 ; Add 17 to both sides

3a = 18 ; Divide both sides by 3

a = 6

6 0
2 years ago
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Wendy wants to run 3 miles. Each lap around the track at her school is mile.
aliina [53]

Answer:

Wendy needs to run 3 laps to run 3 miles.

Hope this helps. :)

7 0
2 years ago
Keisha had $54 when she went to a bookstore. She
storchak [24]
54 - a = 12 i believe
4 0
3 years ago
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Write an equation for a circle with a diameter that has endpoints at (–4, –7) and (–2, –5). Round to the nearest tenth if necess
Zinaida [17]

since we know the endpoints of the circle, we know then that distance from one to another is really the diameter, and half of that is its radius.

we can also find the midpoint of those two endpoints and we'll be landing right on the center of the circle.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-2-(-4)]^2+[-5-(-7)]^2}\implies d=\sqrt{(-2+4)^2+(-5+7)^2} \\\\\\ d=\sqrt{2^2+2^2}\implies d=\sqrt{2\cdot 2^2}\implies d=2\sqrt{2}~\hfill \stackrel{~\hfill radius}{\cfrac{2\sqrt{2}}{2}\implies\boxed{ \sqrt{2}}} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-7})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-5})\qquad \qquad \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{-2-4}{2}~~,~~\cfrac{-5-7}{2} \right)\implies \left( \cfrac{-6}{2}~,~\cfrac{-12}{2} \right)\implies \stackrel{center}{\boxed{(-3,-6)}} \\\\[-0.35em] ~\dotfill

\bf \textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \qquad center~~(\stackrel{-3}{ h},\stackrel{-6}{ k})\qquad \qquad radius=\stackrel{\sqrt{2}}{ r} \\[2em] [x-(-3)]^2+[y-(-6)]^2=(\sqrt{2})^2\implies (x+3)^2+(y+6)^2=2

4 0
3 years ago
Read 2 more answers
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