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zlopas [31]
3 years ago
7

A boy flies a kite with a 100-foot-long string. The angle of elevation of the string is 48°. Find the vertical distance between

the kite and the point at which the boy is holding the string.
Mathematics
1 answer:
NeX [460]3 years ago
4 0
The vertical distance will be given by:
sinθ=opposite/hypotenuse
where:
θ=48°
opposite=h
hypotenuse=100 ft
thus
sin 48=h/100
hence
h=100 sin48
h=74.315 ft

Answer: 74.315 ft
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In Exercise, expand the logarithmic expression.<br> ln x(x + 1)
34kurt

Answer:

dy/dx = 2x+1/x(x+1)

Step-by-step explanation:

y = In x(x+1)

Let u = x(x+1) = x^2 + x

du/dx = 2x + 1

y = In u

dy/du = 1/u

dy/dx = dy/du × du/dx = 1/u × 2x+1 = 2x+1/u = 2x+1/x(x+1)

3 0
3 years ago
Fine the difference between the proud of 26.22 and 3.09 and the sum of 3.507,208,11.5,and 16.712.
zlopas [31]
<span>Fine the difference between the proud of 26.22 and 3.09 and the sum of 3.507,208,11.5,and 16.712.: 
</span>
60.800591885
7 0
3 years ago
Marisol buys 3 notebooks that each cost the same amount and 1 calculator that costs $9. The total cost is $21. How much does eac
adoni [48]

Answer: 4 dollars for each notebook

Step-by-step explanation:

21-9= 12

12/3=4

each notebooks costs 4 dollars

4 0
3 years ago
Read 2 more answers
Find an equation of the line containing the given pair of points (5,2) and (-3,5)
anzhelika [568]
(5,2)(-3,5)
slope = (5 - 2) / (-3 - 5) = -3/8

y = mx + b
slope(m) = -3/8
use either of ur points...(5,2)...x = 5 and y = 2
now we sub and find b, the y int
2 = -3/8(5) + b
2 = - 15/8 + b
2 + 15/8 = b
16/8 + 15/8 = b
31/8 = b

so ur equation is : y = -3/8x + 31/8....or 3x + 8y = 31
3 0
3 years ago
Any volunteer please help
den301095 [7]

Answer:

Part A)

The equation in the point-slope form is:

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Part B)

The graph of the equation is attached below.

Step-by-step explanation:

Part A)

Given

  • The point = (-2, 11)
  • m = 4/3

The point-slope form of the line equation is

y-y_1=m\left(x-x_1\right)

Here, m is the slope and (x₁, y₁) is the point

substituting the values m = 4/3 and the point (-2, 11)  in the point-slope form of the line equation

y-y_1=m\left(x-x_1\right)

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Thus, the equation in the point-slope form is:

y-11=\frac{4}{3}\left(x-\left(-2\right)\right)

Part B)

As we have determined the point-slope form which passes through the point (-2, 11) and has a slope m = 4/3

The graph of the equation is attached below.

5 0
3 years ago
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