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Ksivusya [100]
4 years ago
15

Joshua has a ladder that is 12 ft long. He will lean the ladder against a vertical wall. For safety reasons, he wants the angle

the ladder makes with the ground to be no greater than 75°. Is it possible for Joshua to lean the ladder against the wall so that the top of the ladder is at least 11.8 ft above the ground?
Mathematics
1 answer:
xenn [34]4 years ago
8 0
We first determine the height that the top of the ladder will reach given that the angle to be made is no greater than 75°. In the right triangle formed, the hypotenuse is 12. The trigonometric function that is derived from the scenario is,
                                     x = (sin 75°)(12 ft) = 11.59 ft
Thus, the highest point that the ladder will reach should only be approximately equal to 11.59 ft. 
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Answer:

k =(2,1)

JK = 2

Step-by-step explanation:

Given

J = (0,1) ---- (x_1,y_1)

H = (1,-2) --- (x_2,y_2)

I = (3,-2) --- (x_3,y_3)

See attachment for grid

Solving (a): The coordinates of K

The parallelogram has the following diagonals: IJ and HK

Diagonals bisect one another. So:

Midpoint of IJ = Midpoint of HK

This gives:

\frac{1}{2}(I + J) = \frac{1}{2}(H+K)

\frac{1}{2}(x_3+x_1,y_3+y_1) = \frac{1}{2}(x_2+x,y_2+y)

\frac{1}{2}(3+0,-2+1) = \frac{1}{2}(1+x,-2+y)

\frac{1}{2}(3,-1) = \frac{1}{2}(1+x,-2+y)

Multiply through by 2

(3,-1) = (1+x,-2+y)

By comparison:

1 + x = 3

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Solve for x and y

x = 3 - 1 =2

y = -1 +2 = 1

So, the coordinates of k is:

k =(2,1)

The length of JK is calculated using distance (d) formula

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2

J = (0,1) ---- (x_1,y_1)

k =(2,1) ---- (x_2,y_2)

So:

d = \sqrt{(0 - 2)^2 + (1 - 1)^2

d = \sqrt{(- 2)^2 + (0)^2

d = \sqrt{4 + 0

d = \sqrt{4

d = 2

Hence:

JK = 2

8 0
3 years ago
Question 1
olga_2 [115]

Step-by-step explanation:

(7r+5s)^2

use the pattern

(a+b)^2=a^2+2ab+b^2

(7r+5s)^2=(7r+5s)(7r+5s)

Apply FOIL method to multiply the parenthesis

(7r+5s)^2=(7r+5s)(7r+5s)

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