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Wewaii [24]
2 years ago
8

Find the angle measure indicated to the nearest tenth:

Mathematics
1 answer:
makvit [3.9K]2 years ago
6 0

Step-by-step explanation:

Sin<D = Opposite / Hypotenuse

Opposite - EC

Hyp - DE

Sin<D = EC/DE = x/9

we need x to find <D.

so -->Use pythagorean theorem.

DE^2 = EC^2 + DC^2

DE = 9 DC = 7 EC = ?

EC^2 = DE^2 - DC^2 rearranged.

= 9^2 - 7^2

= 81 - 49

EC^2 = 32 Put both sides under square root.

√(EC^2) = √32

EC = 4√2 ~ 5.65.

We now have X which was representing the unknown side EC.

Sin<D = EC/DE = 5.65/9 = 0.627

To find <D Take the sine inverse of of 0.627.

<D = Arcsin(0.627) = 38.82°.

We now know <D. It's <E's turn.

A right angle triangle has a summation of interior angles of 180°.

thus, <em><D + <C + <E = 180°</em>

38.82° + 90° + <E = 180°

128.82° + <E = 180°

subtract both sides by 128.82°

0 + <E = 180° - 128.82°

<em><E = 51.</em><em>2</em><em>°</em>

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Answer:

f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...

We first compute the n-th derivative of f(x)=\ln(1+2x), note that

f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\

Now, if we compute the n-th derivative at 0 we get

f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!

and so the Maclaurin series for f(x)=ln(1+2x) is given by

f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2  \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}

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Answer:a) P(8 of the players numbers are drawn)=1.3×10^-8

b) P(7 of the players number are drrawn)=3.33×10^-c) P(at least 6 of the players number were drawn)=1.84×10^-4

Step-by-step explanation:

Players has 8 combinations of numbers from 1-40. The outcome S contains all the combinations of 8 out of 40

a) P(8 of the players numbers are drawn)= 1/40/8= 1.3×10^-8

There are one in hundred million chances that the draw numbers are precisely the chosen ones.

b) Number of ways of drawing 78 selected numbers from 1-40=8×(40-7)

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There are approximately 300,000 chances that 7 of the players numbers are chosen

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P(at least 6 players numbers are drawn)=P(all 8 chosen are drawn)+P(7 players numbers drawn)+P(6 chosen are drawn) = 1+ 8 x32/40/8 +[8\6 ×32/2]

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Answer:

Step-by-step explanation:

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<u><em>Answer:</em></u>

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<u><em>Explanation:</em></u>

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We also know that a year has 12 months

<u>This means that:</u>

Craig pays \frac{720}{12}= 60 dollars each month for his telephone services

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Hope this helps :)

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3 years ago
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