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steposvetlana [31]
3 years ago
8

HELP ASAP!!!!!!!!!!!!!! WILL MARK BRAINLIEST PICTURES BELOW!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
harkovskaia [24]3 years ago
4 0

5. 6√3

6.  x = 10√3     y = 30

7. x = 34    y = 17√3

8. 30

9. SinA = 3/5       CosA = 4/5

10. Tan20 = 9/x

Multiply both sides by x to get it on the other side

x(tan20) = 9

Divide 9 by tan20 to get x.

x = 9/tan20

x = 24.7



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Solve for the value of a. .<br> (9a+6°<br> 750
Lilit [14]
As you can see those two angles are supplementary which the sum of two angles = 180°

So,

(9a + 6) + 75 = 180

Subtract 75 on both sides:

9a + 6 = 105

Subtract 6 on those sides:

9a = 99

Divide 9 on both sides:

a = 11


-
8 0
3 years ago
Prove or disprove that the point (√5, 12) is on the circle centered at the origin and containing the point (-13, 0). Show your w
pav-90 [236]

Using the equation of the circle, it is found that since it reaches an identity, the point (√5, 12) is on the circle.

<h3>What is the equation of a circle?</h3>

The equation of a circle of center (x_0, y_0) and radius r is given by:

(x - x_0)^2 + (y - y_0)^2 = r^2

In this problem, the circle is centered at the origin, hence (x_0, y_0) = (0,0).

The circle contains the point (-13,0), hence the radius is found as follows:

x^2 + y^2 = r^2

(-13)^2 + 0^2 = t^2

r^2 = 169

Hence the equation is:

x^2 + y^2 = 169

Then, we test if point (√5, 12) is on the circle:

x^2 + y^2 = 169

(\sqrt{5})^2 + 12^2 = 169

25 + 144 = 169

Which is an identity, hence point (√5, 12) is on the circle.

More can be learned about the equation of a circle at brainly.com/question/24307696

#SPJ1

6 0
3 years ago
9) through: (1, 2), slope = 7
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Y = 7x + b
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tell me the drop down box and i can help

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Read 2 more answers
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