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netineya [11]
4 years ago
15

Find the value of x to the nearest degree.

Mathematics
1 answer:
Vinvika [58]4 years ago
3 0

Answer:  77°

<u>Step-by-step explanation:</u>

In relation to ∠X, 3 is the adjacent side and 13 is the opposite side.

tan\ \theta=\dfrac{opposite}{adjacent}\\\\tan\ x=\dfrac{13}{3}\\\\x = tan^{-1}\bigg(\dfrac{13}{3}\bigg)\\\\.\ =77

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Find the measure of angle2<br><br> A. 90<br> B. 50<br> C. 60<br> D. 30
galina1969 [7]
I think it is C. 60.
You can eliminate A because it is not a right angle, and D because the angle is bigger than the given angle. So it is between 50 and 60, and I hope this is right and the picture makes sense

6 0
3 years ago
Recall the question for a circle with center. (H,k) and Radius r. At what point in the first quadrant does the line with equatio
kap26 [50]

Answer:

the point lies at (1.7, 3.6) in the first quadrant.

Step-by-step explanation:

The formula for a circle with centre (h,k) and radius (r) can be expressed as :

(x-h)² + (y-k)² = r²

Now for the expression of the circle with radius 3 and center (0,1), we have:

(x - 0)² + (y - 1)² = 3²

x² + (y-1)² = 9

replacing y = 1.5x + 1 in the above equation, we have:

x² + (( 1.5x + 1 ) - 1 )² = 9

x² + ( 1.5x +1 - 1)² = 9

x² + (1.5x)² = 9

x² + 2.25x² = 9

3.25x² = 9

x² = 9/3.25

x^2 = \dfrac{9}{\dfrac{13}{4}}

x^2 = {9} \times {\dfrac{4}{13}}

x= \sqrt{{9} \times {\dfrac{4}{13}}}

x= \pm 3 \times \dfrac{2}{\sqrt{13}}}

x= \pm \dfrac{6}{\sqrt{13}}}

x = 1.7 in the positive x - axis

Recall that:

y = 1.5x + 1

y = 1.5 (1.7) +1

y = 2.55 +1

y = 3.55

y = 3.6

Therefore, the point lies at (1.7, 3.6) in the first quadrant.

4 0
3 years ago
determine the average rate of change of the function between the given variables h(x) = x; x=a, x=a+h​
Alecsey [184]

Answer:

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Step-by-step explanation:

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We have that a=1, b=3, f(x)=x^2.

Thus, f(b)−f(a)/b−a=((3))^2−(((1))^2)/3−(1) = 4.

6 0
4 years ago
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I just took the test it's
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6 0
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Plz help me on my last question thanks
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Ok if you would to take away 3 it would increase the mean median and mode 

5 0
4 years ago
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