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madreJ [45]
3 years ago
14

A. 96° b. 132° c. 137° d. 89°

Mathematics
1 answer:
inessss [21]3 years ago
5 0

Answer:

c. 137°

Step-by-step explanation:

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A wire attached to the top of a pole reaches a stake in the ground 20 feet from the foot of the pole and makes an angle of 58° w
Finger [1]
It is a right triangle
cos 58 = 20/length of rope
length of rope = 20/cos 58 degrees
= 37.74 feet
6 0
3 years ago
Can someone please answer. There is one problem. There's a picture. Thank you!
schepotkina [342]
Sin(12) ≈ 0.208

  cos(x) = 0.208
  cos(x) = sin(12)
cos(78) = sin(12)

  cos(12) ≈ 0.978
  cos(68) ≈ 0.375
cos(102) ≈ -0.208
  cos(78) ≈ 0.208

The answer is D.
4 0
4 years ago
Please help as soon as possible
patriot [66]
17.5

5 is the og length. Scale factor is 3.5. So

5(3.5)=17.5
3 0
3 years ago
if the variance of the data values in a population is 256, what is the standard deviation of the data values?
kumpel [21]
The standard deviation is the square root of the variance

so if the variance is 256, then the standard deviation is : sqrt 256 = 16 <==
5 0
4 years ago
If the area of a baseball diamond (shape of a square) is 8100 ft squared, how would you find the distance from 1st base to 3rd b
Readme [11.4K]

Answer:

  • Exact distance = 90\sqrt{2} feet
  • Approximate distance = 127.2792 feet.

===========================================================

Explanation:

The area of the square is 8100 square feet, abbreviated ft^2.

Apply the square root to find the distance from any base to its adjacent counterpart (eg: from 1st to 2nd base). So we get sqrt(8100) = 90 ft as that side distance. Notice that 90*90 = 8100.

If you were to draw a line from 1st base to 3rd base, then you would split the square into two congruent right triangles. Each right triangle is isosceles (the two legs being 90 ft each).

Use the pythagorean theorem to find the hypotenuse.

a^2 + b^2 = c^2\\\\c = \sqrt{a^2+b^2}\\\\c = \sqrt{90^2+90^2}\\\\c = \sqrt{2*90^2}\\\\c = \sqrt{2}*\sqrt{90^2}\\\\c = \sqrt{2}*90\\\\c = 90\sqrt{2} \ \text{ ... exact distance}\\\\c \approx 127.2792 \ \text{ ... approximate distance}\\\\

The distance from 1st to 3rd base is roughly 127.2792 feet.

5 0
2 years ago
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