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mylen [45]
2 years ago
9

Find sin (b) in the triangle

Mathematics
2 answers:
Gala2k [10]2 years ago
5 0

Answer:

sinβ = 15/17

General Formulas and Concepts:

<u>Math</u>

  • Simplifying

<u>Trigonometry</u>

  • [Right Triangles Only] SOHCAHTOA
  • [Right Triangles Only] sinθ = opposite over hypotenuse

Step-by-step explanation:

<u>Step 1: Define</u>

Angle θ = β

Opposite Leg = 15

Hypotenuse Leg = 17

<u>Step 2: Find</u>

  1. Substitute in variables [sine]:                                                                          sinβ = 15/17
iogann1982 [59]2 years ago
5 0

Answer: 15/17

Step-by-step explanation: Khan Academy

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I hope this helps you




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5 0
3 years ago
James took a math quiz in which he earned 1 point for every correct answer and lost 0.5 points for every incorrect answer.James
Nuetrik [128]

Answer:

9

Step-by-step explanation:

6x0.5=3-12=9

4 0
3 years ago
In 2000 the total amount of gamma ray bursts was recorded at 6.4 million for a city. In 2005, the same survey was made and the t
ratelena [41]

Let's assume

It started in 2000

so, t=0 in 2000

P_0=6.4million

we can use formula

P(t)=P_0 e^{rt}

we can plug value

P(t)=6.4 e^{rt}

In 2005, the same survey was made and the total amount of gamma ray bursts was 7.3 million

so, at t=2005-2000=5

P(t)=7.3 million

we can plug value and then we can solve for r

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now, we can plug back

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now, we have

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4 0
3 years ago
Consider the function f given by f(x)=x*(e^(-x^2)) for all real numbers x.
NISA [10]

Answer:

\frac{\sqrt{\pi}}{4}

Step-by-step explanation:

You are going to integrate the following function:

g(x)=x*f(x)=x*xe^{-x^2}=x^2e^{-x^2}  (1)

furthermore, you know that:

\int_0^{\infty}e^{-x^2}=\frac{\sqrt{\pi}}{2}

lets call to this integral, the integral Io.

for a general form of I you have In:

I_n=\int_0^{\infty}x^ne^{-ax^2}dx

furthermore you use the fact that:

I_n=-\frac{\partial I_{n-2}}{\partial a}

by using this last expression in an iterative way you obtain the following:

\int_0^{\infty}x^{2s}e^{-ax^2}dx=\frac{(2s-1)!!}{2^{s+1}a^s}\sqrt{\frac{\pi}{a}} (2)

with n=2s a even number

for s=1 you have n=2, that is, the function g(x). By using the equation (2) (with a = 1) you finally obtain:

\int_0^{\infty}x^2e^{-x^2}dx=\frac{(2(1)-1)!}{2^{1+1}(1^1)}\sqrt{\pi}=\frac{\sqrt{\pi}}{4}

5 0
3 years ago
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Kobotan [32]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
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