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BabaBlast [244]
3 years ago
11

Find the coordinates of the midpoint HX giving that H(-1,3) and X(7,-1)

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
8 0
D is the answer for the question
BigorU [14]3 years ago
6 0
Midpoint is : (x1 + x2) / 2 , (y1 + y2) / 2
(-1,3)....x1 = -1 and y1 = 3
(7,-1)...x2 = 7 and y2 = -1
now we sub and solve
m = (-1 + 7)/2 , (3 - 1) / 2
m = (6/2, 2/2)
m = (3,1) <===
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yan [13]
<span>Rate of pump A: 1/8 of a pool per hour 
Rate of pump B: 1/9 of a pool per hour 
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So if they work together, the two pumps have a combined rate of 25/72 of a pool per hour (i.e in one hour, the two pumps will empty 25/72 of the pool) 

</span><span>But we want to empty ONE pool (not 25/72 of one). So we need to multiply 25/72 by some number x to get 1.
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<span>Now solve for x 
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7 0
3 years ago
Find the value of tan(sin^-1(1/2))
suter [353]

If you know that \sin\dfrac\pi3=\dfrac12, then you know right away

\tan\left(\sin^{-1}\dfrac12\right)=\tan\dfrac\pi3=\dfrac1{\sqrt}3=\dfrac{\sqrt3}3

###

Otherwise, you can derive the same result. Let \theta=\sin^{-1}\dfrac12, so that \sin\theta=\dfrac12. \sin^{-1} is bounded, so we know -\dfrac\pi2\le\theta\le\dfrac\pi2. For these values of \theta, we always have \cos\theta\ge0.

So, recalling the Pythagorean theorem, we find

\cos^2\theta+\sin^2\theta=1\implies\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\dfrac12\right)^2}=\dfrac{\sqrt3}2

Then

\tan\theta=\tan\left(\sin^{-1}\dfrac12\right)=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}=\dfrac{\sqrt3}3

as expected.

5 0
3 years ago
Read 2 more answers
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Oliga [24]
First you have to find a common denominator which would be 24.
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5 0
3 years ago
Read 2 more answers
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Ber [7]

Answer:

C, the 3rd one

Hope this helped! Have a great day!

3 0
3 years ago
An interval has the notation (-6,-2). find the midpoint
notka56 [123]
<span>(6+8)/2 = 14/2 = 7 hope this helps

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8 0
3 years ago
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