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jonny [76]
3 years ago
9

Solve for x 90° \12x-22) 170°

Mathematics
1 answer:
Y_Kistochka [10]3 years ago
5 0

Answer:

x = 6

Step-by-step explanation:

Using this theorem to get the other angles of the triangle.

Angle at the vertex is equal to half of the angle at the intercepted arc

For the arc with angle 170degrees

Angle at the vertex = 1/2 * 170

Angle at the vertex = 85 degrees

For the arc with angle 90degrees

Angle at the vertex = 1/2 * 90

Angle at the vertex = 45 degrees

Taking the sum of the 3 angles and equating to 180

85 + 45 + 12x - 22 = 180

130-22+12x = 180

12x = 180 - 108

12x = 72

x = 72/12

x = 6

Hence the value of x is 6

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Step-by-step explanation:

Find the Radius so 9/2= 4.5

Area = π r²

Area 3.14(20.25)

Area equles 63.62

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3 years ago
How do you prove the Pythagorean Theorem with Similar Triangles using either a two-column, paragraph, or flow chart proof?
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3 years ago
Instructions require you to dilute a powder in 3 quarts of water. You only have a container that is marked in liters. How many l
Julli [10]

Answer:

1 quart = .947 L

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3 years ago
Find four consecutive integers such that the sum of the two largest subtracted from three times the sum of the two smallest is 7
serious [3.7K]

Answer:

18, 19, 20, 21

Step-by-step explanation:

Just like any of these problems, we should start by forming an equation for us to get a reference and plug in. We'll be using x as our variables.

As we have four consecutive integers (and not multiples) we can assume that the integers will be x, x+1, x+2, and x+3.

The sum of the two largest integers we have equals: n+2 + n+3 = 2n+5

and three times the sum of the two smallest = 3(n + n+1) = 6n+3

and the sum of t he two largest subtracted from three times the sum of the two smallest = (6n+3) - (2n+5) = 4n-2

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7 0
3 years ago
Integrate sin²2x cos²2x dx
DochEvi [55]
\sin^2x=\dfrac{1-\cos2x}2
\cos^2x=\dfrac{1+\cos2x}2

From the identities above, you have

\sin^22x\cos^22x=\dfrac{(1-\cos4x)(1+\cos4x)}4=\dfrac{1-\cos^24x}4

Applying once more, you have

\dfrac{1-\cos^24x}4=\dfrac{1-\dfrac{1+\cos8x}2}4=\dfrac{1-\cos8x}8

So,

\displaystyle\int\sin^22x\cos^22x\,\mathrm dx=\frac18\int(1-\cos8x)\,\mathrm dx=\frac x8-\frac1{64}\sin8x+C
7 0
3 years ago
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