A pair of designer sunglasses was marked up 10% and is now selling for $165.00. What was the original price of the sunglasses?
2 answers:
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Answer:
(a) 4 seconds
Step-by-step explanation:
The given point (4, 0) on the parabola tells you the answer to the question: the coin was in the air 4 seconds before its height is 0. It is no longer in the air at that point.
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<em>Additional comment</em>
In order to match the rest of the problem statement, the first given x-intercept must be (-2, 0).
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>
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>