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Xelga [282]
3 years ago
12

A rectangle Has an area of 105 ft.² if the sum of the length and width is 26 feet find the dimensions include units

Mathematics
1 answer:
zhenek [66]3 years ago
5 0
LW=105
L+W=26
L=26-W
(26-W)W=105
26W-W^2=105
0=W^2-26W+105
-105=W^2-26W
-105+169=W^2-26W+169
64= (W-13)^2
8= -(W-13) and (W-13)
8= -W+13
5=W

8= W-13
21=W

105= 5*21

Answer: 5 feet x 21 feet
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The coordinate of point N after the dilation is N' = (-8/3, 8/3)

<h3>Dilation of coordinates</h3>

Dilation is a transformation technique for changing the size of an object using a scale factor.

Given the coordinate of N from the figure as (-4, 4). If this coordinate point is dilated by a factor of 2/3, then the resulting coordinate will be:

N' = (2/3(-4), 2/3(4))

N' = (-8/3, 8/3)

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<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

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To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

\implies 3x + y = -7

\implies3x + (3x - 1) = -7

\implies3x + 3x - 1 = -7

Combine like terms as needed;

\implies 3x + 3x - 1 = -7

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Add 1 to both sides of the equation;

\implies 6x - 1 + 1 = -7 + 1

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Divide 6 to both sides of the equation;

\implies \dfrac{6x}{6}  = \dfrac{-6}{6}

\implies x = -1

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

\implies y = 3(-1) - 1

\implies     \ \ = -3 - 1

\implies     = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.

\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.

\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

\implies 0 = (6x) + (6)

\implies0 = 6x + 6

This problem is now an algebraic problem. Isolate "x" to determine its value.

\implies 0 - 6 = 6x + 6 - 6

\implies -6 = 6x

\implies -1 = x

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

\implies y = 3(-1) - 1

\implies y = -3 - 1

\implies y = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

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<h3>How to determine the coordinates of the partition?</h3>

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