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Georgia [21]
3 years ago
10

The length of a rectangle is 10 yd less than three times the width, and the area of the rectangle is 77 yd^2. Find the dimension

s of the rectangle.
Mathematics
1 answer:
Vitek1552 [10]3 years ago
8 0

Answer:

W=7 and L=11

Step-by-step explanation:

We have two unknowns so we must create two equations.

First the problem states that  length of a rectangle is 10 yd less than three times the width so: L= 3w-10

Next we are given the area so: L X W = 77

Then solve for the variable algebraically. It is just a system of equations.

3W^2 - 10W - 77 = 0

(3W + 11)(W - 7) = 0

W = -11/3 and/or W=7

Discard the negative solution as the width of the rectangle cannot be less then 0.

So W=7

Plug that into the first equation.

3(7)-10= 11 so L=11

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2 years ago
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I don’t know how to solve this
V125BC [204]

Assuming that both triangles are an exact copy of one another, it is safe to assume that 3y-7 is equal to 41. Set up an equation

3y-7=41

Add 7 to both sides

3y=48

Divide both sides by 3

y=16


Now to find PN.

Based on what we know, we can assume that MP = PN. Let's make some equations!

MP = 17x-8     PN = 11x+4

17x-8 = 11x+4

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Add 8 to both sides

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Divide by 2

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Substitute 2 in for x in the equation for PN

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Multiply 11 by 2

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