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djverab [1.8K]
3 years ago
14

If these two figures are similar, what is the measure of the missing angle? 180 90 70 110

Mathematics
2 answers:
yawa3891 [41]3 years ago
8 0

Answer:

70

Step-by-step explanation:

motikmotik3 years ago
3 0

Answer:

70

Step-by-step explanation:

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Please Help ASAP. Will Mark as Brainliest
marin [14]

Answer:

B

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What is the answer 5(x-4)=3x+4
ruslelena [56]

Step-by-step explanation:

Simplify the left side:

5x - 20 = 3x + 4

Move all terms containing  x  to the left side of the equation:

2x - 20 = 4

Move all terms not containing  x  to the right side of the equation:

2x = 24

Divide each term by  2  and simplify:

x = 12

4 0
3 years ago
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What are the missing sides and angles of this triangle? What is the area?
Butoxors [25]
This does not appear to be a right triangle.  However, we know 2 sides and the included angle, so can find the unknown side length.  Let x represent this length.  Then:

x^2 = (9 m)^2 + (12 m)^2 - 2(9m)(12 m)*cos 30 degrees, or

x^2 = 81 + 144 - 216(sqrt(3) / 2).  Please solve for x^2 and then solve the result for x, making sure to choose the positive value.  The result will be the length of the side opposite the 30 degree angle.

With 1 of 3 angles known, and 3 of 3 sides known, you can use the Law of Sines to find the other two angles.  As a reminder, the Law of Sines looks like this:

   a            b              c
-------- = --------- = ----------
sin A       sin B       sin C.

You can give the 30-deg angle any name you want; then a, the length of the side opposite the 30-deg angle, which you have just found.  And so on.  
4 0
3 years ago
State if each pair of ratios forms a proportion 12/24 and 3/4
raketka [301]

Answer:

Step-by-step explanation:

12/24 reduces to 1/2, whereas 3/4 does not reduce.  This pair of ratios does not form a proportion.

If both ratios reduced to the same common value, that would represent a proportion.  

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3 years ago
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Find f. f ″(x) = x^−2, x > 0, f(1) = 0, f(6) = 0
marin [14]

If you do in fact mean f(1)=f(6)=0 (as opposed to one of these being the derivative of f at some point), then integrating twice gives

f''(x) = -\dfrac1{x^2}

f'(x) = \displaystyle -\int \frac{dx}{x^2} = \frac1x + C_1

f(x) = \displaystyle \int \left(\frac1x + C_1\right) \, dx = \ln|x| + C_1x + C_2

From the initial conditions, we find

f(1) = \ln|1| + C_1 + C_2 = 0 \implies C_1 + C_2 = 0

f(6) = \ln|6| + 6C_1 + C_2 = 0 \implies 6C_1 + C_2 = -\ln(6)

Eliminating C_2, we get

(C_1 + C_2) - (6C_1 + C_2) = 0 - (-\ln(6))

-5C_1 = \ln(6)

C_1 = -\dfrac{\ln(6)}5 = -\ln\left(\sqrt[5]{6}\right) \implies C_2 = \ln\left(\sqrt[5]{6}\right)

Then

\boxed{f(x) = \ln|x| - \ln\left(\sqrt[5]{6}\right)\,x + \ln\left(\sqrt[5]{6}\right)}

3 0
2 years ago
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