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BigorU [14]
2 years ago
12

In the following diagram, QR is tangent to circle P. Which of the following could be the missing side lengths? Select all that a

pply. (3 answers)
A. 8 and 4\sqrt{3}
B. 2 and 4\sqrt{2}
C. 3 and 5
D. 2 and 2\sqrt{5}
E. 6 and 10
Please show your work too

Mathematics
1 answer:
timofeeve [1]2 years ago
5 0

Answer:

QR is tangent to circle P at point Q

=> RQP = 90 deg

=> RQP is right triangle.

=> Applying Pythagorean theorem for right triangle RQP:

    QP^2 + QR^2 = RP^2

=>4^2 + QR^2 = RP^2

=> Option A is correct.

(4^2 + (4*\sqrt{3})^2 = 16 + 48 = 64 = 8^2)

=> Option C is correct.

(4^2 + 3^2 = 16 + 9 = 25 = 5^2)

=> Option D is correct

(4^2 + 2^2 = 16 + 4 = 20 = (2\sqrt{5})^2)

Hope this helps!

:)

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Answer:

The standard deviation of car age is 2.17 years.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 7.5

(a) If 99.7% of the ages are between 1 year and 14 years, what is the standard deviation of car age?

This means that 1 is 3 standard deviations below the mean and 14 is 3 standard deviations above the mean.

So

14 = 7.5 + 3\sigma

I want to find \sigma

3\sigma = 6.5

\sigma = \frac{6.5}{3}

\sigma = 2.17

The standard deviation of car age is 2.17 years.

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Eli had $10 but he lost some of it. He mom doubled the money he had left. Eli wrote the expression 2(10-k) how much money he has
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Answer:

Step-by-step explanation:

According to the first expression:

2(10-k): where,

2:shows the double amount of money which ELI has

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k:shows the amount of money which is lost.

10-k:shows the amount of money which ELI has after losing some amount.

According to the second expression:

20-2k where,

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2k:s hows the twice of amount which is lost

20-2k: shows the amount of money which is left with ELI after his mom gave him some money....

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (
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(a) Suppose h_n=r^n is a solution for this recurrence, with r\neq0. Then

r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}
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So we expect a general solution of the form

h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n

With h_0=0,h_1=1,h_2=1,h_3=2, we get four equations in four unknowns:

\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}

So the particular solution to the recurrence is

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(b) Let G(x)=\displaystyle\sum_{n\ge0}h_nx^n be the generating function for h_n. Multiply both sides of the recurrence by x^n and sum over all n\ge4.

\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n
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