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matrenka [14]
3 years ago
10

The area of a regular octagon is 25 square cm . What is the area of a regular octagon with sides five times as large as the side

s of the first octagon?
Mathematics
1 answer:
Nutka1998 [239]3 years ago
3 0
When you change a dimension, it will change EACH dimension in the measurement you are finding.

Area is a two dimensional measurement, so you will need to multiply each dimension by 5, which is the same as multiplying the area by 25.

So, 25 square cm x 5 x 5= 625 square cm.

The area of the new OCTAGON is 625 square cm.
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A(x) = P(1.02)^x
A(x) = P(1 + 0.02)^x
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<img src="https://tex.z-dn.net/?f=4x%20%2B%203y%20%3D%200%20%5C%5C5y%20%2B%2053%20%3D%2011x%20%5C%5C%20" id="TexFormula1" title=
Anvisha [2.4K]

Answer:

x = 3, y = -4

Step-by-step explanation by substitution:

Solve the following system:

{4 x + 3 y = 0 | (equation 1)

5 y + 53 = 11 x | (equation 2)

Express the system in standard form:

{4 x + 3 y = 0 | (equation 1)

-(11 x) + 5 y = -53 | (equation 2)

Swap equation 1 with equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

4 x + 3 y = 0 | (equation 2)

Add 4/11 × (equation 1) to equation 2:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+(53 y)/11 = -212/11 | (equation 2)

Multiply equation 2 by 11/53:

{-(11 x) + 5 y = -53 | (equation 1)

0 x+y = -4 | (equation 2)

Subtract 5 × (equation 2) from equation 1:

{-(11 x)+0 y = -33 | (equation 1)

0 x+y = -4 | (equation 2)

Divide equation 1 by -11:

{x+0 y = 3 | (equation 1)

0 x+y = -4 | (equation 2)

Collect results:

Answer: {x = 3 , y = -4

_____________________________________

Solve the following system:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Choose an equation and a variable to solve for.

In the first equation, look to solve for x:

{4 x + 3 y = 0

5 y + 53 = 11 x

Hint: | Isolate terms with x to the left hand side.

Subtract 3 y from both sides:

{4 x = -3 y

5 y + 53 = 11 x

Hint: | Solve for x.

Divide both sides by 4:

{x = -(3 y)/4

5 y + 53 = 11 x

Hint: | Perform a substitution.

Substitute x = -(3 y)/4 into the second equation:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Choose an equation and a variable to solve for.

In the second equation, look to solve for y:

{x = -(3 y)/4

5 y + 53 = -(33 y)/4

Hint: | Isolate y to the left hand side.

Subtract 53 - (33 y)/4 from both sides:

{x = -(3 y)/4

(53 y)/4 = -53

Hint: | Solve for y.

Multiply both sides by 4/53:

{x = -(3 y)/4

y = -4

Hint: | Perform a back substitution.

Substitute y = -4 into the first equation:

Answer: {x = 3 , y = -4

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Simplify the following expression 65l/13q
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Answer:

\dfrac{5l}{q}

Step-by-step explanation:

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Answer:

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There are 2 zeros so there are 2 factors

( x- -4) ( x-8) = 0

(x+4) (x-8) =0

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