<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>of</em><em> </em><em>option</em><em> </em><em>D</em><em>.</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em>⤴</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>
Answer:
Does the answer help you?
Answer:
(-4, -2)
Step-by-step explanation:
The x-coordinate would stay the same, but the y-coordinate would be halved. Thus, the corresponding point would be (-4, -2).
Answer:
k < 50$
Step-by-step explanation:
She wants the keyboard for less than 50$, so the price, k, is less than 50 in this inequality.
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.
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