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3 years ago

Someone help me I have to turn in tomorrow

Mathematics

1 answer:

drek231 [11]3 years ago

4 0

I can’t read it? It’s too far from the camera.

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Question 9<br> The triangles shown are

yaroslaw [1]

Answer:

The triangles shown are<u> proportional</u>

Step-by-step explanation:

Two parallel lines are cut by a tranversal so the opposite and corresponding angles are congruent.

B = N

M = M

12 x 2 = 24 , So the sides of the triangle are proportional meaning they are parallel.

Side MN would be 12

Side MP would be 16

6 0

3 years ago

What do i do <br> its at 4:00 pm today

inn [45]

Answer:

I cant see

Explanation:

Attach it properly?

7 0

2 years ago

PLEASE HELP INEQUALITYS!!

kolezko [41]

Same i am sorry need point

5 0

3 years ago

What’s the correct answer for this?

joja [24]

Answer:

MN = 48

Step-by-step explanation:

Since AB is a diameter , so it divides MN into 2 equal parts i.e. MO = NO

7x-4 = 6x

7x-6x = 4

x = 4

Now

MN = MO + NO

MN = 7x-4+6x

MN = 7(4)-4+6(4)

MN = 28-4+24

MN = 48

6 0

3 years ago

Read 2 more answers

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:

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

Then add all these boxed integrated together then we'll get:

$\displaystyle{4\left(\tan x - 2x + \dfrac{1}{4}\sin 2x + \dfrac{1}{2} x\right) + C}$

Expand 4 in the expression:

$\displaystyle{4\tan x - 8x +\sin 2x + 2 x + C}\\\\\displaystyle{4\tan x + \sin 2x - 6x + C}$

Therefore, the answer is:

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

4 0

1 year ago