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DIA [1.3K]
3 years ago
5

The length of a rectangle is 5 Inches more than its width, x. The area of a rectangle can be represented by the equation

Mathematics
1 answer:
Paul [167]3 years ago
4 0

Answer:

Factor:

x^2+5x-300=0

(x-15)(x+20)=0

x=15 is the width

x+5=20 is the length

Step-by-step explanation:

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FinnZ [79.3K]

Answer:

3 ÷ 0.5

Step-by-step explanation:

Since each large square is 1 whole, and theirs five out of 10 columns that are distinctivly a different orange, it'd be 3 divided by 0.5.

5 0
2 years ago
Please calculate this limit <br>please help me​
Tasya [4]

Answer:

We want to find:

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}

Here we can use Stirling's approximation, which says that for large values of n, we get:

n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} =  \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}

Now we can just simplify this, so we get:

\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\

And we can rewrite it as:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}

7 0
3 years ago
What single transformation was applied to quadrilateral A to get quadrilateral B?
Tems11 [23]

Answer: translation.

8 0
3 years ago
Kierston needs to paint a wall of the youth center. She knows that 1 can of paint covers an area of 2.2 square meters. Kierston
Step2247 [10]

Answer:

See Explanation

Step-by-step explanation:

Given

1\ can = 2.2m^2

Required

Determine the number of cans for the wall

The dimension of the wall is not given. So, I will use the following assumed values:

Length=20m

Width = 44m

First, calculate the area of the wall

Area = Length * Width

Area = 20m * 44m

Area = 880m^2

If 1\ can = 2.2m^2

Then x = 880m^2

Cross Multiply:

x * 2.2m^2 = 1 * 880m^2

x * 2.2m^2 = 880m^2

x * 2.2 = 880

Make x the subject

x = \frac{880}{2.2}

x = 400

400 cans using the assume dimensions.

So, all you need to to is, get the original values and follow the same steps

4 0
3 years ago
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