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Ann [662]
3 years ago
14

Each side of a square is increasing at a rate of 3 cm/s. at what rate is the area of the square increasing when the area of the

square is 16 cm2?
Mathematics
1 answer:
deff fn [24]3 years ago
3 0
The squares length is increasing by 5.33 cm/ s
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Which property was used to solve the given inequality?
Amiraneli [1.4K]

Subtraction Property of Inequality is used to solve the given inequality

<em><u>Solution:</u></em>

Given inequality is:

y + 8 < - 8

To solve the inequality, subtract 8 on both sides,

We know that,

Adding or subtracting the same quantity from both sides of an inequality leaves the inequality symbol unchanged

Also, by Subtraction Property of Inequality,

If you minus a number from one side of an inequality, you have to minus that same number from the other side of the inequality as well

y + 8 < - 8\\\\\text{Subtract 8 from both sides of inequality }\\\\y + 8 - 8 < -8  - 8\\\\simplify,\\\\y

Thus, Subtraction Property of Inequality is used

4 0
3 years ago
Can someone help me?
SOVA2 [1]

Answer:

what does the picture say?

Step-by-step explanation:

7 0
3 years ago
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Hector took out a small loan of $900 for 15 months. The simple interest rate on the loan was 15%. How much will he pay in intere
Liula [17]
So you have to find 15 percent of 900 which would be 135
3 0
3 years ago
The value of 5 in 506712 is how many times the value of 5 in 324859
Oxana [17]
It is 1.55 times the value of the second number. 
8 0
3 years ago
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The scatter plot shows a correlation between the years and the rainfall in centimeters in Tennessee.
nataly862011 [7]

Answer:

7

4

Step-by-step explanation:

The <u>actual values</u> are shown on the given graph as <u>blue points</u>.

The <u>line of regression</u> is shown on the given graph as the <u>red line</u>.

From inspection of the graph, in the year 2000 the actual rainfall was 43 cm, shown by point (2000, 43).  It appears that the regression line is at y = 50 when x is the year 2000.

⇒ Difference = 50 - 43 = 7 cm

<u>In 2000, the actual rainfall was </u><u>7</u><u> centimeters below what the model predicts</u>.

From inspection of the graph, in the year 2003 the actual rainfall was 44 cm,  shown by point (2003, 40).  It appears that the regression line is at y = 40 when x is the year 2003.

⇒ Difference = 44 - 40 = 4 cm

<u>In 2003, the actual rainfall was </u><u>4</u><u> centimeters above what the model predicts.</u>

3 0
2 years ago
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