Answer: the length is 11 cm.
The width is 7 cm.
Step-by-step explanation:
Let L represent the length of the rectangular plastic box.
Let W represent the width of the rectangular plastic box.
The area of the rectangular top of the box is 77 square cm. This means that
LW = 77- - - - - - - ;- - - -1
The plastic box has a length 4 cm longer than its width. This means that
L = W + 4
Substituting L = W + 4 into equation 1, it becomes
(W + 4)W = 77
W² + 4W = 77
W² + 4W - 77 = 0
W² + 11W - 7W - 77 = 0
W(W + 11) - 7(W + 11) = 0
W - 7 = 0 or W + 11 = 0
W = 7 or W = - 11
Since the width cannot be negative, then W = 7cm
L = 77/7 = 11 cm
Answer: The flagpole is about 22.3 feet tall.
This is roughly the size of a two story building, assuming each floor is 10 feet or so.
Since this height is under 25 ft, this flagpole is in compliance with the regulation.
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Work Shown:
The horizontal leg of the triangle is 36 ft. This is the adjacent leg to the reference angle 25 degrees.
The vertical leg of the triangle is x-5.5, and this leg is the opposite side to the reference angle 25 degrees.
Use the tangent rule to connect the opposite and adjacent sides.
Solve for x.
tan(angle) = opposite/adjacent
tan(25) = (x-5.5)/36
36*tan(25) = x-5.5
x-5.5 = 36*tan(25)
x = 36*tan(25)+5.5
x = 22.28707569358
x = 22.3 ft is the approximate height of the flagpole
Make sure your calculator is in degree mode.
The height is not over 25 ft, so this flagpole meets the requirements.
15b is the answer to this question hope this was helpful
Answer:
It would be D because X^2 is different from 3x you would just multiply X^2 by 1 and 3x by 1 to get x^2+3x