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melisa1 [442]
3 years ago
9

Write an equation in slope-intercept form for the line that passes through the given points and is (a) parallel to the given lin

e and (b) perpendicular to the given line. 1. (-7,3);x=4
2.(5,-1);y=4x-7
Please help!!

Mathematics
1 answer:
murzikaleks [220]3 years ago
3 0
Remark
This is costing you an awful lot of points, but it is a little hard to read. If I'm not interpreting it correctly, leave some sort of note below my answer.

Question One
In the first question you want a line that is parallel to x = 4 and goes through the point (-7,3). If I'm reading this correctly then the line you want is x = -7. You only need the x value of the given point. I'll put a graph there to show you what it looks like on a grid.  The graph is the left one for this question.

Question Two
Here I think you want the line going through (5,-1) and perpendicular to y = 4x - 7. I will make a graph for that one as well.

Begin with the slope
Two lines are perpendicular to each other if their slopes multiply to - 1
Given slope (m1) = 4
perpendicular slope = m2

Formula
m1 * m2 = - 1
4 * m2 = -1      Divide by 4
m2 = -1/4

So far what you have is
y = (-1/4)x + b

Now use the given point to solve for b
x = 5
y = - 1
b = ??

-1 = (-1/4)*5 + b
-1 = -5/4 + b  Add 5/4 to both sides
-1 + 5/4 + b   Make -1 into -4/4
b = -4/4 + 5/4
b = (-4 + 5)/4
b = 1/4

So the line that you want is
y = (-1/4) x + 1/4

Answers
One:  x = - 7
Two: y = (-1/4)x + 1/4

Note
The right graph is of y = 4x - 7 and y = (-1/4)x + 1/4

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3 years ago
One of the tallest buildings in a country is topped by a high antenna. The angle of elevation from the position of a surveyor on
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Answer:

a. distance of the surveyor to the base of the building = 2051.90 ft

b. height of the building = 1384 ft

c. Angle of elevation from the surveyor to the top of the antenna = 38.31°

d. Height of antenna  =  237.08 ft

Step-by-step explanation:

​The picture above is a illustration of the described event.

a = the height of the flag

b = the height of the building

c = distance of the surveyor from the base of the building

the angle of elevation from the position of the surveyor on the ground to the top of the building = 34°  

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distance from her position to the top of the flag  = 2615 ft

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using the SOHCAHTOA principle

cos 34° = c/2475

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c = 2051.8679921

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(b) How tall is the​ building

The height of the building = b

sin 34° = opposite /hypotenuse

0.5591929035 = b/2475

b =  0.5591929035  × 2475

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let the angle = ∅

cos ∅ = adjacent/hypotenuse

cos ∅ = 2051.90/2615

cos ∅ =  0.784665392

∅ = cos-1  0.784665392

∅ =   38.310258303

∅ =  38.31°

​(d) How tall is the​ antenna?

height of the antenna = a

sin 38.31° = opposite/hypotenuse

sin 38.31° = (a + b)/2615

sin 38.31° × 2615 = (a + b)

(a + b) =  0.6199159917  × 2615

(a + b) =  1621.0803182

(a + b) = 1621. 08 ft

Height of antenna = 1621. 08 - 1384.00  =  237.08031822 ft

Height of antenna  =  237.08 ft

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