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KonstantinChe [14]
3 years ago
10

How do you do number 1?Whenever I tried to answer it, I always get fraction. help me.​

Mathematics
1 answer:
Flura [38]3 years ago
6 0

Answer:

The pairs are (13,15) and (-15,-13).

Step-by-step explanation:

If n is an odd integer, the very next odd integer will be n+2.

n+1 is even (so we aren't using this number)

The sum of the squares of (n) and (n+2) is 394.

This means

(n)^2+(n+2)^2=394

n^2+(n+2)(n+2)=394

n^2+n^2+4n+4=394               since (a+b)(a+b)=a^2+2ab+b^2

Combine like terms:

2n^2+4n+4=394

Subtract 394 on both sides:

2n^2+4n-390=0

Divide both sides by 2:

n^2+2n-195=0

Now we need to find two numbers that multiply to be -195 and add up to be 2.

15 and -13 since 15(-13)=-195 and 15+(-13)=2

So the factored form is

(n+15)(n-13)=0

This means we have n+15=0 and n-13=0 to solve.

n+15=0

Subtract 15 on both sides:

n=-15

n-13=0

Add 13 on both sides:

n=13

So if n=13 , then n+2=15.

If n=-15, then n+2=-13.

Let's check both results

(n,n+2)=(13,15)

13^2+15^2=169+225=394.  So (13,15) looks good!

(n,n+2)=(-15,-13)

(-15)^2+(-13)^2=225+169=394.  So (-15,-13) looks good!

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hichkok12 [17]

Answer:

Width of the rectangular banner is 3\frac{5}{6} ft.

Step-by-step explanation:

Area of the rectangular banner = 2\frac{7}{8} square feet

                                                    = \frac{23}{8} square feet

Area of rectangle is given by,

Area = Length × Width

Length of the banner = \frac{3}{4} feet

From the formula,

\frac{23}{8}=\frac{3}{4}\times W

W = \frac{\frac{23}{8} }{\frac{3}{4} }

    = \frac{23}{8}\times \frac{4}{3}

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Therefore, width of the rectangular banner is 3\frac{5}{6} ft

7 0
3 years ago
Write as a product: 25a2−0.01b10
Yakvenalex [24]

Answer:

We conclude that:

25a^2-0.01b^{10}=\left(5a+0.1b^5\right)\left(5a-0.1b^5\right)

Step-by-step explanation:

Given the expression

25a^2-0.01b^{10}

Rewrite 25a^2-0.01b^{10} as \left(5a\right)^2-\left(\sqrt{0.01}b^5\right)^2

so

25a^2-0.01b^{10}=\left(5a\right)^2-\left(\sqrt{0.01}b^5\right)^2

Apply difference of two squares formulas:

x^2-y^2=\left(x+y\right)\left(x-y\right)

so

\left(5a\right)^2-\left(\sqrt{0.01}b^5\right)^2=\left(5a+\sqrt{0.01}b^5\right)\left(5a-\sqrt{0.01}b^5\right)

                               =\left(5a+0.1b^5\right)\left(5a-0.1b^5\right)    ∵ \sqrt{0.01}=0.1,

Therefore, we conclude that:

25a^2-0.01b^{10}=\left(5a+0.1b^5\right)\left(5a-0.1b^5\right)

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