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The length of a rectangle is 1 inch less than twice its width, and its perimeter is 40 inches. Find the width of the rectangle.


7 in.
13 in.
14 in.




If the second equation is multiplied by 5 and then the equations are added, the result is _____.

x + 5y = 3
2x - y = -2


9x = 13
11x = -7
9x = -7


For the following system, use the second equation to make a substitution for x in the first equation.

x + 3y = 5
x + 7 = 2y

What is the resulting equation?


2y + 2x + 3y = 5
2y + 7 + 3y = 5
2y - 7 + 3y = 5




Graph the following system of equations. Click on the graph until the correct one appears.

y > -x + 1
y ≤ 2x - 3


Find A B if A = {3, 6, 9, 12} and B = {2, 4, 6, 8, 10}.



{6}
{2, 3, 4, 6, 8, 9, 10, 12}


Choose the correct mathematical translation for the given statement.

She is at most twenty years old.


x 20
x ≥ 20

70 POINTS can someone help me with this i rlly need it The length of a rectangle is 1 inch less than twice its width and its perimeter is 40 inches Find the wid class=

Respuesta :

Answer:

1) 7 in

2) 11x=-7

3)2y - 7 + 3y = 5

4) i DONT'KNOWN

Step-by-step explanation:

1)

step 1

the length of the rectangle is 1 inch less than twice its width ( in other words , you have to subtract  1 to twice the width to get the length)

Let

L represents the length

W represent the width

Hence

Twice width= 2w

Now , replace

L=2w-1 equation (1)

Also

The perimeter =40 inches

[tex]perimeter= 2(L+W) 40=2(L+W) equation (2)[/tex]

Step 2

Solve for w

Replace equation (1) in equation (2)

[tex]40=2(L+W) equation (2)\\40=2((2W-1)+W)\\40=2(3W-1)\\40=6W-2\\Add 2 in both sides\\40+2=6W-2+2\\42=6W\\Divide both sides by 6\\\frac{42}{6}=\frac{6W}{6} \\7=W[/tex]

so ,the width is 7 in.

2) Given the equations

[tex]x+5y=3\\2x-y=-2[/tex]

The  first step is to multiply the second equation by 5 (both sides of the equation)

[tex]5(2x-y=-2)\\5(2x-y)=5.(-2)\\5.2x-5.y=-10\\10x-5y=-10\\[/tex]

Next is to add both equations

x+5=3 +10x-5y=-10 = 11x+0y=-7

The result is 11x=-7

3)[tex]x+3y=5\\x+7=2y[/tex]

To make a substitution for x using the second equation in the first equation you :

1. solve for x the second equation

Subtract 7 in both sides of the equation:

[tex]x+7-7=2y-7\\x=2y-7\\[/tex]

2.Substitute the x in the first equation for the value of x you get in step 1 :

[tex]2y-7+3y=5[/tex]

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