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vampirchik [111]
3 years ago
14

Quick!!

Mathematics
1 answer:
Illusion [34]3 years ago
6 0

Answer:

The height of the building is approximately;

B. 154 ft

Step-by-step explanation:

From the given drawing of the right triangle formed by the measure of the angle to the top of the building, 64°, the height of the building, 'h', and their distance from the bottom of the building, 75 ft., we have;

The reference angle of the right triangle = The given 64° angle

The adjacent leg length to the reference angle = 75 ft.

The opposite leg length to the reference angle  The height of the building

From the given information, the trigonometric ratio we can use to find the height is the tangent of the reference angle, given as follows;

tan (Reference \, angle) = \dfrac{Opposite \ leg \ length}{Adjacent \ leg\ length}

Therefore, we have;

tan (64^{\circ}) = \dfrac{The \ height \ of \ the \ building, h}{75 \, ft.}

The height of the building, h = 75 ft. × tan(64°) = 153 feet 9\dfrac{9}{32} inches

Therefore, the height of the building ≈ 154 feet.

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How do you simplify (-5x^5+14)-(11x^2+1+11x^5)
lord [1]
(-5x ⁵+14)-(11x ²+1+11x ⁵)
Like terms: -5x ⁵-11x ⁵ = -16x ⁵
Like terms also: 14-1 = 13
Simplified all together: -16x ⁵ - 11x ²+ 13
(Always put highest degree first)
6 0
3 years ago
A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a
Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

                                   p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

p^4 (1-p)

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

(1-p)^4p

Therefore the probability that you stop at the fifth flip would be

                                    p^4 (1-p)  + (1-p)^4 p

(b)

The expected numbers of flips needed would be

\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1}  = 1/p

Therefore, suppose that  p = 0.5, then the expected number of flips needed would be 1/0.5  = 2.

7 0
3 years ago
Round 6.85565 to the nearest tenth !
emmasim [6.3K]

Answer:

It is 6.9

Step-by-step explanation:

6.85565

put

6.85

5 can be rounded up so

6.9 is answer

3 0
4 years ago
1.what is a sports car average acceleration if it can go from 0m/s to 27m/s in 6.0 seconds?
Flauer [41]

Answer:

1.4.5m/s^2

2.6.25m/s^2

3.15m/s^2

4.0.05m/s^2

Step-by-step explanation:

a=v-u

----

t

27-0

------ = 27/6

6

=4.5m/s^2

u=4.5m/s

v=24.5m/s

t=3.2s

(24.5-4.5)÷3.2

=20/3.2

=6.25m/s^2

v=80m/s

u=50m/s

t=2s

(80-50)÷2

15m/s^2

v=0.80m/s

u=0.50m/s

t=6s

(0.80-0.50)÷6

=0.05m/s^2

5 0
2 years ago
Ty bought 5 folders that cost $3 each. How much did he spend on folders?
Novay_Z [31]
5 x 3 = 15

Ty spent $15 on binders
3 0
3 years ago
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