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3 years ago

15 POINTS!!! Solve the problem. Type the number in the box that correctly completes the sentence.

Mathematics

1 answer:

OverLord2011 [107]3 years ago

7 0

Answer:

5 Miles

Step-by-step explanation:

Given, Height of the tower is 3 miles and the distance from the base is 4 miles.

According the Pythagoras Theorem, the square of the length of the hypotenuse is equal to the sum of squares of the lengths of other two sides of the right-angled triangle.

So, putting the data in Pythagoras theorem, let's consider A= 3 and B= 4.

So C will be,

sqrt(A^2 + B^2 )= C

Or, sqrt( 3^2+ 4^2)= C

Or, sqrt( 9+ 16) = C

Or, sqrt(25) = C

Hence, C = 5

So, The distance would be 5 Miles.

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Given a line segment that contains the points A,B, & C in order,if AB = 5x + 4, BC = 4x - 8, and AC = 23, find the length of

Ivenika [448]

Answer:

Step-by-step explanation:

AB + BC = AC , that is

5x + 4 + 4x - 8 = 23

9x - 4 = 23 ( add 4 to both sides )

9x = 27 ( divide both sides by 9 )

x = 3

Then

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3 years ago

What is the value of x in the equation 1/3x + 7 = -2

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The value of x is -27

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4 years ago

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3 years ago

A turtle is crawling up a hill that rises 6 feet for every horizontal change of 36 feet. Write the slope of the hill as a

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3 years ago

Use the given point on the terminal side of angle θ to find the value of the trigonometric function indicated.

Fofino [41]

Answer: 19) 117° 20) 53°

21) 229° 22) 119°

23) 155° 24) 323°

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

Use Pythagorean Theorem to find r: x² + y² = r²

19) x = $-\sqrt{17}$ , y = 8, r = 9

$\cos \theta=\dfrac{x}{r} \rightarrow \quad \cos \theta =\dfrac{-\sqrt{17}}{9} \rightarrow \quad \theta = cos^{-1}\bigg(\dfrac{-\sqrt{17}}{9}\bigg)\rightarrow \quad \theta = \large\boxed{117^o}$

20) x = 3, y = 4, r = 5

$\cos \theta=\dfrac{x}{r} \rightarrow \quad \cos \theta =\dfrac{3}{5} \rightarrow \quad \theta = cos^{-1}\bigg(\dfrac{3}{5}\bigg)\rightarrow \quad \theta = \large\boxed{53^o}$

21) x = $-\sqrt7$ , y = -3, r = 4

$\sin \theta=\dfrac{y}{r} \rightarrow \quad \sin \theta =\dfrac{-3}{4} \rightarrow \quad \theta = sin^{-1}\bigg(\dfrac{-3}{4}\bigg)\rightarrow \quad \theta = \large\boxed{229^o}$

22) x = $-\sqrt{15}$ , y = 7, r = 8

$\tan \theta=\dfrac{y}{x} \rightarrow \quad \tan \theta =\dfrac{7}{-\sqrt{15}} \rightarrow \quad \theta = tan^{-1}\bigg(\dfrac{7}{-\sqrt{15}}\bigg)\rightarrow \quad \theta = \large\boxed{119^o}$

23) x = -13, y = 6, r = $\sqrt{205}$

$\tan \theta=\dfrac{y}{x} \rightarrow \quad \tan \theta =\dfrac{6}{-13} \rightarrow \quad \theta = tan^{-1}\bigg(\dfrac{6}{-13}\bigg)\rightarrow \quad \theta = \large\boxed{155^o}$

24) x = 4, y = -3, r = 5

$\sin \theta=\dfrac{y}{r} \rightarrow \quad \sin \theta =\dfrac{-3}{5} \rightarrow \quad \theta = sin^{-1}\bigg(\dfrac{-3}{5}\bigg)\rightarrow \quad \theta = \large\boxed{323^o}$

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3 years ago