$\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}}
{\stackrel{slope}{-\cfrac{1}{9}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{9}{1}}\qquad \stackrel{negative~reciprocal}{+\cfrac{9}{1}}\implies 9}$

4 0

3 years ago

Read 2 more answers

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ale4655 [162]

Answer:

$Area = 58 \ units^2$

Step-by-step explanation:

We can start by finding the third side of the triangle/the side of the square using the Pythagorean Theorem:

$a^2+b^2=c^2$

In this case the side length of the square would be represented by the variable "c" as it is the hypotenuse:

$3^2+7^2=c^2$

$9+49=c^2$

$c^2=58$

$c=\sqrt{58}$

Since the area of a square is the side length square then...

$Area =( \sqrt{58} )^2$

The square root and the squared cancel out giving us...

$Area = 58 \ units^2$

3 0

2 years ago

Factor the following problems -13=-4a-2b+c 3=4a+2b+c 5=16a+4b+c Please Help!

Pie

1) -13 = -4a - 2b + c
-13 = -2(2a + b) + c

2) 3 = 4a + 2b +c
3 = 2(2a + b) + c

3) 5 = 16a + 4b + c
5 = 4(4a + b) + c

[ Final Answers are in bold ]

Hope this helps!

8 0

2 years ago

Korvikt [17]

Step-by-step explanation:

As you have done till

3p + 8 = 20

3p = 20 - 8

3p = 12

p = 12 / 3 = 4

5 0

3 years ago