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lutik1710 [3]
3 years ago
8

The length of a rectangle is 15 and its width is w. The perimeter of the rectangle is, at most, 50. Which inequality can be used

to find the longest possible width?
1) 30 + 2w < 50
2) 30 + 2w ≤ 50
3) 30 + 2w > 50
4) 30 + 2w ≥ 50
Mathematics
1 answer:
Vladimir79 [104]3 years ago
5 0

Answer:

2) 30 + 2w ≤ 50

Step-by-step explanation:

First, let's find the perimeter of the rectangle in terms of w. We have 2*(15+w) = 30+2w because the perimeter of a rectangle can be written as 2*(length + width). The problem says that the perimeter is at most 50. This means that it cannot be greater than 50, so it would have to be less than or equal to 50. It can be equal to 50 because that is what at most means. Putting this information into an inequality, we get 30 + 2w ≤ 50.

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Answer:

Option C is correct, i.e. 2x² +7x -4.

Step-by-step explanation:

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3 0
3 years ago
Need step by step detailed answer
kolezko [41]

Answer:

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Step-by-step explanation:

If we have two lines in slope intercept form as

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In other words, m_2 = -\frac{1}{m_1}

We have the first line as

y = \frac{1}{4}x + 5\\\\

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So slope of perpendicular line is -4 which implies

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It is always a good idea to plot these graphs and see if they fit the data provided

The attached plot shows the two graphs and you can see they are perpendicular to each other and the perpendicular line(the answer) passes through point (-3,8)

4 0
1 year ago
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Answer:

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Step-by-step explanation:

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olchik [2.2K]

Answer:

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Step-by-step explanation:

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