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Advocard [28]
2 years ago
13

To cover A rectangle region of her yard penny needs atleast 170.5 square feet of sod. The length of the region is 15.5 feet. Wha

t are the possible widths of the region? Write an inequality and solve. I know the answer is 11, but how would you write the inequality?
Mathematics
1 answer:
stiks02 [169]2 years ago
7 0
The area needed is at least 170.5 ft².

Let w  =  the width of the rectangular area.
Because the length is 15.5 ft, therefore the calculated area should be at least 170.5 ft². That is
(15.5 ft)*(w ft) ≥ 170.5 ft²
w ≥ 170.5/15.5
w ≥ 11 ft

Answer:  w ≥ 11 ft
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Step-by-step explanation:

2x+3≤x−5

Step 1: Subtract x from both sides.

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Step 2: Subtract 3 from both sides.

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4. The cube root parent function is vertically stretched by a factor of 3, reflected in the x-axis,
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Using translation concepts, it is found that the equation that represents the new function is:

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The cube root parent function is given by:

y = \sqrt[3]{x}

<h3>Translation:</h3>

The function goes through these following translations:

  • Vertically stretched by a factor of 3, that is, it is multiplied by 3, hence y = 3\sqrt[3]{x}.
  • Reflected over the x-axis, that is, it is multiplied by -1, hence y = -3\sqrt[3]{x}.
  • Translated 8 units down, that is, 8 is subtracted from the function, hence y = -3\sqrt[3]{x} - 8.

Then, the equation is:

  • y = -3\sqrt[3]{x} - 8

To learn more about translation concepts, you can take a look at brainly.com/question/13671886

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During each trial, 8 events could have occurred:


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The plot of points obtained by Eric shows the frequency in which each of these 8 events occurred during the 53 trials.


The fact that the point diagram shows only one point on event 8, means that it occurred only once during the 53 trials, and compared to the other 7 events, this was the one with a lower frequency (the one that obtained the the highest frequency was the 4-tailed one, with 11 points).


This means that it is very unlikely to get 8 queues when casting 8 coins.


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