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Ber [7]
3 years ago
9

a rectangle pool is 9 ft wide. the pool has an area of 117 square feet. what is the perimeter of the pool?

Mathematics
2 answers:
Softa [21]3 years ago
6 0
9*13=117
(9*2)+(13*2)=44
44 feet
SashulF [63]3 years ago
3 0
You need to find the length: 117 ÷ 9 =13 ft
Find the perimeter: (13+9) ×2 = 44 ft
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2(3 + 4x) = 8x + 6<br> How Many solutions?
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Answer:

Infinite number of solutions

Step-by-step explanation:

2(3 + 4x) = 8x + 6\\6 + 8x = 8x+6\\8x-8x=6-6\\0=0

For any value of x, both equations are equal to each other because both sides are identical

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Let X = the number of nonzero digits in a randomly selected 4-digit PIN that has no restriction on the digits. What are the poss
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Answer:

X = {0, 1, 2, 3, 4}

Step-by-step explanation:

If X is the number of nonzero digits in a 4-digit PIN with no restriction on the digits, the pin can have up to 4 nonzero digits.

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If the PIN is 0000 then X = 4

If the PIN is 2045 then X = 3

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Please please answer this correctly I have to finish this today as soon as possible
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Step-by-step explanation:

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Next go half a box to the right and two boxes up, and plot your point there. that is (.5,2).

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n
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Answer:

(-1,1),(4,-2)

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point (x,y)

Distance between points (x_1,y_1),(x_2,y_2) is given by \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]

Put (x,y)=(-1,1)

34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34

which is true. So, (-1,1) can be a vertex

Put (x,y)=(4,-2)

34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34

which is true. So, (4,-2) can be a vertex

Put (x,y)=(1,1)

34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22

which is not true. So, (1,1) cannot be a vertex

Put (x,y)=(2,-2)

34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22

which is not true. So, (2,-2) cannot be a vertex

Put (x,y)=(4,-1)

34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30

which is not true. So, (4,-1) cannot be a vertex

Put (x,y)=(-1,4)

34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70

which is not true. So, (-1,4) cannot be a vertex

So, possible points for the vertex are (-1,1),(4,-2)

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