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3 years ago

What is the angle measure of x in the trapezoid below?

Mathematics

2 answers:

patriot [66]3 years ago

6 0

Answer:

i put 106 and i got it right

Step-by-step explanation:

lorasvet [3.4K]3 years ago

4 0

106 because the inside of a trapezoid has to equal 360

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Simplify x = cos[2 arcsin(3/5)]. -7/25 7/25 4/5

Marina CMI [18]

Answer:

7/25

Step-by-step explanation:

Let $\theta= \arcsin(\frac{3}{5})$ so we have $x=\cos(2\theta)$

As $\cos(2\theta)=\cos^2\theta-\sin^2\theta$ , we'll have $\cos[2\arcsin(\frac{3}{5})]=\bigr[\cos(\arcsin(\frac{3}{5}))\bigr]^2-\bigr[(\sin(\arcsin(\frac{3}{5}))\bigr]^2$

To determine $\cos(\arcsin(\frac{3}{5}))$ , construct a right triangle with an opposite side of 3 and a hypotenuse of 5. This is because since $\theta=\arcsin(\frac{3}{5})$ , then $\sin\theta=\frac{3}{5}=\frac{\text{Opposite}}{\text{Hypotenuse}}$ . If you recognize the Pythagorean Triple 3-4-5, you can figure out that the adjacent side is 4, and thus, $\cos\theta=\frac{4}{5}=\frac{\text{Adjacent}}{\text{Hypotenuse}}$ . This means that $\cos(\arcsin(\frac{3}{5}))=\frac{4}{5}$ .

Hence, $\cos[2\arcsin(\frac{3}{5})]=(\frac{4}{5})^2-(\frac{3}{5})^2=\frac{16}{25}-\frac{9}{25}=\frac{7}{25}$

5 0

2 years ago

Are they all right? If not please tell me what the answers are. Thanks! (Math)

MAXImum [283]

Answer:

yes

Step-by-step explanation:

They are absolutely correct. Good Job

4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%20%5Cinfty%20%20%5Cfrac%7B%20%5Csqrt%5B%20%20%5Cscriptsize%5Cphi%

Rasek [7]

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

$I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx$

Replace $x \to x^{\frac1\phi} = x^{\phi-1}$ :

$I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

Split the integral at x = 1. For the integral over [1, ∞), substitute $x \to \frac1x$ :

$\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

The integrals involving tan⁻¹ disappear, and we're left with

$I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}$

8 0

2 years ago

Find cos in the triangle There are 2 pictures

Elanso [62]

Answer is c

something that helps me is
soh- opposite hypotenuse cah - adjacent and hypotenuse toa- opposite and adjacent
soh cah toa

7 0

3 years ago

Describe the motion of a particle with position (x, y) as t varies in the given interval. (For each answer, enter an ordered pai

DerKrebs [107]

Answer:

The motion of the particle describes an ellipse.

Step-by-step explanation:

The characteristics of the motion of the particle is derived by eliminating $t$ in the parametric expressions. Since both expressions are based on trigonometric functions, we proceed to use the following trigonometric identity: