Jeremy has 50 ft of fencing to enclose a rectangular-pen for his dog. He plans on using a side of his house for one side of the
1 answer:
The dimensions such as length is 18 and width is 16 or length is 32 and width is 9 will produce an enclosed area of 288ft.
<h3>What is the dimension of length and width?</h3>We know that,
P = l + 2w
50 = l + 2w
50 - 2w = l
A = l x w
288 = (50 - 2w)w
288 = 50w -2w²
2w² - 50w + 288
By dividing throughout by 2
w² - 25w + 144
This is in quadratic equation form.
By applying quadratic formula,
w = (-(-25) ± ) / 2
w = (-(-25) ± 7)/2
w1 = (-(-25) + 7)/2
w1 = (25 + 7) / 2
w1 = 16
w2 = (-(-25) - 7)/2
w2 = (25 - 7) / 2
w2 = 9
By substituting the values in eq1,
A = l x w
if w = 16
288 = l x 16
l = 288/16
l = 18
if w = 9
288 = l x 9
l = 288/9
l = 32
The dimensions will be length is 18 and width is 16 or length is 32 and width is 9.
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The range of a function is all values of y the function can have. The range of this function is all positive numbers, including zero:
In order to graph f(x) + 2, we just need to translate the graph 2 units up. To find the new points, we need to increase all y-coordinates by 2:
(-2, 6), (-1, 3), (0, 2), (1, 3), (2, 6)
Domain: {x ∈ ℝ}
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Then, in order to graph f(x) - 2, we just need to translate the graph 2 units down. To find the new points, we need to decrease all y-coordinates by 2:
(-2, 2), (-1, -1), (0, -2), (1, -1), (2, 2)
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Range: {y ∈ ℝ | y ≥ -2}
