Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by

The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>
2/8 and 4/16 are equivalent to 1/4
Answer:
ok loll
Step-by-step explanation:
Answer:
- C. No. Vinay also multiplied the right-hand side of the equation by --5, which is incorrect.
Step-by-step explanation:
<u>Proper application of the property:</u>
- - 5(m - 2) - 25 = -5m - 5(-2) - 25 = -5m + 10 - 25 = -5m - 15
Vinay multiplied 5 and 25 as well, which is incorrect.
Correct answer choice is C.
Answer:
x-y=4
and
x+y=32 is <u>your</u><u> </u><u>equation</u>
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<u>x</u><u>-</u><u>y</u><u>=</u><u>4</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>(</u><u>1</u><u>)</u>
<u>x</u><u>+</u><u>y</u><u>=</u><u>3</u><u>2</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>.</u><u>(</u><u>2</u><u>)</u>
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<u>2</u><u>x</u><u>=</u><u>3</u><u>6</u>
<u>x</u><u>=</u><u>3</u><u>6</u><u>/</u><u>2</u><u>=</u><u>1</u><u>6</u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>
<u>s</u><u>u</u><u>b</u><u>s</u><u>t</u><u>i</u><u>t</u><u>u</u><u>t</u><u>i</u><u>n</u><u>g</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>y</u><u> </u><u>in</u><u> </u><u>equation</u><u> </u><u>1</u>
<u>1</u><u>6</u><u>-</u><u>y</u><u>=</u><u>4</u>
<u>1</u><u>6</u><u>-</u><u>4</u><u>=</u><u>y</u>
<u>y</u><u>=</u><u>1</u><u>2</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u>