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yuradex [85]
3 years ago
10

The perimeter of a parallelogram must be no less than 40 feet. The length of the rectangle is 6 feet. What are the possible meas

urements of the width? Write an inequality to represent this problem. Use w to represent the width of the parallelogram. [Hint: The formula for finding the perimeter of a parallelogram is P = 2 l + 2 w . What is the smallest possible measurement of the width? Justify your answer by showing all your work.
Mathematics
1 answer:
tatuchka [14]3 years ago
5 0

Answer: 14\ ft

Step-by-step explanation:

Given

Length of rectangle is 6\ ft

Perimeter must be greater than 40 ft

Suppose l and w be the length and width of the rectangle

\Rightarrow \text{Perimeter P=}2(l+w)\\\Rightarrow P\geq 40\\\Rightarrow 2(l+w)\geq40\\\Rightarrow l+w\geq20\\\Rightarrow w\geq20-6\\\Rightarrow w\geq14\ ft

So, the smallest width can be 14\ ft

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valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in y=ax^2+bx+c

it's easier to find vertex form first then expand to get general form

vertex form is y=a(x-h)^2+k where the vertex is (h,k) and a is a constant


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vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

h_1=a(t-10)^2+50

to find the value of a, subsitute another point

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h_1=\frac{-1}{2}(t^2-20t+100)+50

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same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is h_2=a(t-10)^2+72

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