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2 years ago

What is m _BEC OB) 70° OA) 65° OD) 120° OC) 50°

Mathematics

1 answer:

just olya [345]2 years ago

5 0

Answer:

C) 50°

Step-by-step explanation:

$m\angle BEC= 180\degree-(65\degree+65\degree)$

$\implies m\angle BEC= 180\degree-130\degree$

$\implies m\angle BEC= 50\degree$

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If x^2=y^2+z^2 what does x equal?

Zepler [3.9K]

Answer:

$\displaystyle x = \sqrt{y^2 + z^2}$

General Formulas and Concepts:

Pre-Algebra

Equality Property

Algebra i

Terms/Coefficients

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle x^2 = y^2 + z^2$

Step 2: Solve for x

[Equality Property] Square root both sides: $\displaystyle x = \sqrt{y^2 + z^2}$

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2 years ago

Solve the right triangle

tresset_1 [31]

SR =sin (angle) = opposite leg/ hypotenuse

sin(52) = SR/ 7.6

SR = 7.6 x sin(52)

SR = 5.988 ( round answer as needed.)

QR = cos(angle) = adjacent leg / hypotenuse

Cos(52) = QR/ 7.6

QR = 7.6 x cos(52)

QR = 4.679. ( round answer as needed).

Angle S = 180 - 90 -52 = 38 degrees

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2 years ago

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3 years ago

Evaluate the integral following

alina1380 [7]

Answer:

$\displaystyle{4\tan x + \sin 2x - 6x + C}$

Step-by-step explanation:

We are given the integral of:

$\displaystyle{\int 4(\sec x - \cos x)^2 \, dx}$

First, we can use a property to separate a constant out of integrand:

$\displaystyle{4 \int (\sec x - \cos x)^2 \, dx}$

Next, expand the expression (integrand):

$\displaystyle{4 \int \sec^2 x - 2\sec x \cos x + \cos^2 x \, dx}$

Since $\displaystyle{\sec x = \dfrac{1}{\cos x}}$ then it can be simplified to:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2\dfrac{1}{\cos x} \cos x + \cos^2 x \, dx}\\\\\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \cos^2 x \, dx}$

Recall the formula:

$\displaystyle{\int \dfrac{1}{\cos ^2 x} \, dx = \int \sec ^2 x \, dx = \tan x + C}\\\\\displaystyle{\int A \, dx = Ax + C \ \ \tt{(A \ and \ C \ are \ constant.)}$

For $\displaystyle{\cos ^2 x}$ , we need to convert to another identity since the integrand does not have a default or specific integration formula. We know that:

$\displaystyle{2\cos^2 x -1 = \cos2x}$

We can solve for $\displaystyle{\cos ^2x}$ which is:

$\displaystyle{2\cos^2 x = \cos2x+1}\\\\\displaystyle{\cos^2x = \dfrac{\cos 2x +1}{2}}$

Therefore, we can write new integral as:

$\displaystyle{4 \int \dfrac{1}{\cos^2 x} - 2 + \dfrac{\cos2x +1}{2} \, dx}$

Evaluate each integral, applying the integration formula:

$\displaystyle{\int \dfrac{1}{\cos^2x} \, dx = \boxed{\tan x + C}}\\\\\displaystyle{\int -2 \, dx = \boxed{-2x + C}}\\\\\displaystyle{\int \dfrac{\cos 2x +1}{2} \, dx = \dfrac{1}{2}\int \cos 2x +1 \, dx}\\\\\displaystyle{= \dfrac{1}{2}\left(\dfrac{1}{2}\sin 2x + x\right) + C}\\\\\displaystyle{= \boxed{\dfrac{1}{4}\sin 2x + \dfrac{1}{2}x + C}}$