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2 years ago

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An astronomer wishes to measure on a photograph the distance between the image of a certain star and three other star images nea

rby. If eight images are nearby, how many choices of the three stars does he have?

2 answers:

jeka942 years ago

8 0

Answer: 56

Step-by-step explanation:

Given: An astronomer wishes to measure on a photograph the distance between the image of a certain star and three other star images nearby.

The total number of images nearby = 8

The number of images chosen by astronomer = 3

Then, the number of choices of the three stars without being in order is given by :-

$^8C_3=\frac{8!}{3!(8-3)!}=\frac{8\times7\times6\times5!}{3!5!}=56$

Hence, he has 56 choices .

Alexeev081 [22]2 years ago

4 0

8 C 3 = 8! / (3!(8 - 3)!) = 8! / (3! * 5!) = (8 * 7 * 6 * 5!) / (3! * 5!) = (8 * 7 * 6) / (6) = 56

<span>56 choices. </span>

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The pressure at sea level is 111 atmosphere and increases at a constant rate as depth increases. When Sydney dives to a depth of

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Answer:

we have P(x) = mx + 1

Step-by-step explanation:

Allow me to revise your question for a better understanding:

<em>The pressure at sea level is 1 atmosphere and increases at a constant rate as depth increases. When Sydney dives to a depth of 23 meters, the pressure around her is 3.3 point, 3 atmospheres. The pressure p in atmospheres is a function of x, the depth in meters.</em>

My answer:

Given:

At O meter the the pressure is 1 (0, 1)

At 23 meters the the pressure is 3.3 (23, 3.3)

From that, we can form a linear equation with the standard form:

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Substitute the point (0, 1) into (1) we have:

1 = 0.1*0 + b

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2 years ago

How do I solve w(x)=4x 5; find w(-8)?

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Plug -8 in for x: 4(-8) + 5 solve: -32 + 5 -27

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Answer:

Option 1:(x-4)^2+y^2=100

Step-by-step explanation:

Given center = (h,k) = (4,0)

The point (-2,8) lies on circle which means the distance between the point and center will be equal to the radius.

So,

The distance formula will be used:

$d = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} \\=\sqrt{(4+2)^{2}+(0-8)^{2}}\\=\sqrt{(6)^{2}+(-8)^{2}}\\=\sqrt{36+64}\\ =\sqrt{100}\\ =10\ units$

Hence radius is 10.

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Putting the values

(x-4)^2+(y-0)^2=10^2

(x-4)^2+y^2=100

Hence option 1 is correct ..

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with critical points at $t$ such that

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$3(\cos^3t-3\cos t\sin^2t)+4\sin t\cos t=0$

$\cos t(3\cos^2t-9\sin^2t+4\sin t)=0$

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So either

$\cos t=0\implies t=\dfrac{(2n+1)\pi}2$

or

$12\sin^2t-4\sin t-3=0\implies\sin t=\dfrac{1\pm\sqrt{10}}6\implies t=\sin^{-1}\dfrac{1\pm\sqrt{10}}6+2n\pi$

where $n$ is any integer. We get 8 solutions over the given interval with $n=0,1,2$ from the first set of solutions, $n=0,1$ from the set of solutions where $\sin t=\dfrac{1+\sqrt{10}}6$ , and $n=1$ from the set of solutions where $\sin t=\dfrac{1-\sqrt{10}}6$ . They are approximately

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