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Alex787 [66]
3 years ago
7

A skateboard ramp at a park has an inclination of 54º and its base is 12 feet long. What is the length of the ramp?​

Mathematics
1 answer:
Brilliant_brown [7]3 years ago
5 0

Answer:  

the ans is in the picture with the steps  

(hope it helps can i plz have brainlist :D hehe)

Step-by-step explanation:

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3=x+9<br> PLEASE SHOW WORK. SOLVE AND CHECK.
Vika [28.1K]

Answer:

x = —6

Step-by-step explanation:

3 = x + 9

Collect like terms

3 — 9 = x

— 6 = x

x = —6

4 0
3 years ago
Consider the procedure used below to solve the given equation.
muminat
The first mistake was made in step 1 because they took out the x’s it should be 10x-25+2x+8=43
5 0
3 years ago
Please help!!! ill give brainliest
yawa3891 [41]

Answer:

Step-by-step explanation:

divide the number then add on

7 0
3 years ago
Read 2 more answers
The length of a rectangle is 7 more than the width the area is 744 square centimeters find length and width of rectangle
andrezito [222]

Answer:

l=31\ cm\\\\w=24\ cm

Step-by-step explanation:

The formula that is used to calculate the area of a rectangle is:

A=lw

Where "l" is the lenght and "w" is the width.

You know that the area of that rectangle is:

A=744\ cm^2

And, according to the exercise, its lenght is 7 more than its width; then:

l=w+7

Then, you can make the corresponding substitution into the formula A=lw:

 744=(w+7)w

Simplify:

744=w^2+7w\\\\w^2+7w-744=0

Factor the equation. Find two numbers whose sum is 7 and whose product is -744. These are 31 and -24.

Then, you get:

(w-24)(w+31)=0\\\\w_1=24\\\\w_2=-31

The width of the rectangle is the positive value:

w=24\ cm

Then, the lenght is:

l=24+7\\\\l=31\ cm

8 0
3 years ago
PLZ HELP...Enter in standard form the equation of the line passing through the given point and having the given slope.
Alenkinab [10]

Standard form for the equation of the line is

$y=-\frac{2}{3} x-\frac{25}{3}

Solution:

Given point is (2, –5).

slope of the line m = -\frac{2}{3}

Here, x_1=2, y_1=-5

Equation of a line passing through the point:

y-y_1=m(x-x_1)

$y-(-5)=-\frac{2}{3} (x-(-5))

$y+5=-\frac{2}{3} (x+5)

$y+5=-\frac{2}{3} x-\frac{10}{3}

Subtract 15 from both sides of the equation.

$y+5-5=-\frac{2}{3} x-\frac{10}{3}-5

$y=-\frac{2}{3} x-\frac{25}{3}

Standard form for the equation of the line is

$y=-\frac{2}{3} x-\frac{25}{3}

7 0
3 years ago
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