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Harlamova29_29 [7]
3 years ago
5

Write 3 ratios equal to 4/36 ?

Mathematics
2 answers:
MrMuchimi3 years ago
8 0
This is similar to fractions.

1:9
2:18
3:27
Elena-2011 [213]3 years ago
4 0
<span>2/18
1/9
3/27
...............................</span>
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F(x) = 3x - 3 <br> g(x) = 4x+5 <br> find (f•g) (3) <br><br> only numerical answers work
Andreas93 [3]

Answer:

48 bcoz u do 4(3)+5 do f(g) than for f(17) u do 3(17)-3

6 0
3 years ago
AJ is the bisector of ∠HJI. Determine the value of x.
atroni [7]

Answer:

C

Step-by-step explanation:

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Evaluate the following integral using trigonometric substitution
serg [7]

Answer:

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

Step-by-step explanation:

We are given the following integral:

\int \frac{dx}{\sqrt{9-x^2}}

Trigonometric substitution:

We have the term in the following format: a^2 - x^2, in which a = 3.

In this case, the substitution is given by:

x = a\sin{\theta}

So

dx = a\cos{\theta}d\theta

In this question:

a = 3

x = 3\sin{\theta}

dx = 3\cos{\theta}d\theta

So

\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}

We have the following trigonometric identity:

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

So

1 - \sin^{2}{\theta} = \cos^{2}{\theta}

Replacing into the integral:

\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C

Coming back to x:

We have that:

x = 3\sin{\theta}

So

\sin{\theta} = \frac{x}{3}

Applying the arcsine(inverse sine) function to both sides, we get that:

\theta = \arcsin{(\frac{x}{3})}

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

8 0
3 years ago
Please help I would love that!
Kazeer [188]

Answer:

Area= 7234.56

Step-by-step explanation:

area= πr²

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so that means radius is 48

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