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Pepsi [2]
3 years ago
5

f a storm window has an area of 400 square inches, what are the dimensions if the window is 12 inches wider than it is high (w x

h)?
Mathematics
2 answers:
anzhelika [568]3 years ago
5 0
Well if the window is 40 x 10  the L is 52 and the w 22 
denis23 [38]3 years ago
4 0
A=xy and we are told y=12+x so

A=12x+x^2 and A=400 so

x^2+12x=400

x^2+12x-400=0

x=(-12±√1744)/2

x≈14.88

So the dimensions are approximately 14.88" by 26.88"
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700+a=3944-b=c-80=3000​
velikii [3]

Answer:

A = 3244

B = 864

C = 3080

Step-by-step explanation:

 700 + a = 3944

 -700          -700

a = 3244

 3944 - b = c - 80 = 3000

c - 80 = 3000

  +80       +80

c = 3080

3944 - b = c ---> 3944 - b = 3080

                          -3944        -3944

                               -b = -864 --->  \frac{-b}{-1} = \frac{-864}{-1}

7 0
3 years ago
Use mental math to find a solution for 2 over eight= 3over x+2
-Dominant- [34]

The Answer: x= −12/7

5 0
3 years ago
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−7m+4&lt;−45 <br><br> its an one-step problem help me please
aleksandr82 [10.1K]

Answer:

x > 7

Step-by-step explanation:

- 7m + 4 < - 45 ( subtract 4 from both sides )

- 7m < - 49

Divide both sides by - 7 , reversing the symbol as a result of dividing by a negative quantity.

m > 7

7 0
2 years ago
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The sides of a triangle have lengths 4, 9, and 11. What kind of triangle is it?
aalyn [17]

Answer:Scalene Triangle

Step-by-step explanation:

In a scalene triangle,all sides are not equal

4 0
3 years ago
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A company manufactures televisions. The average weight of the televisions is 5 pounds with a standard deviation of 0.1 pound. As
Semenov [28]

Answer:

0.2564\text{ pounds}

Step-by-step explanation:

The 90th percentile of a normally distributed curve occurs at 1.282 standard deviations. Similarly, the 10th percentile of a normally distributed curve occurs at -1.282 standard deviations.

To find the X percentile for the television weights, use the formula:

X=\mu +k\sigma, where \mu is the average of the set, k is some constant relevant to the percentile you're finding, and \sigma is one standard deviation.

As I mentioned previously, 90th percentile occurs at 1.282 standard deviations. The average of the set and one standard deviation is already given. Substitute \mu=5, k=1.282, and \sigma=0.1:

X=5+(1.282)(0.1)=5.1282

Therefore, the 90th percentile weight is 5.1282 pounds.

Repeat the process for calculating the 10th percentile weight:

X=5+(-1.282)(0.1)=4.8718

The difference between these two weights is 5.1282-4.8718=\boxed{0.2564\text{ pounds}}.

8 0
3 years ago
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