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nata0808 [166]
3 years ago
8

Write four different expressions for the perimeter of a pentagon whose sides are all s-2 units long?

Mathematics
1 answer:
Masteriza [31]3 years ago
4 0

Hey there!

1. 5(s-2) This is because s-2 is the length of each side, and there are 5 sides in a pentagon all of that length, so multiplying it by 5 like this shows its perimeter.

2. 5s-10 This can be found by distributing the first equation, multiplying both s and -2 by 5.

3. (s-2) + (s-2) + (s-2) + (s-2) + (s-2) Each grouping in the parenthesis represents each of the 5 sides, so adding them all together will get you the perimeter.

4. 5(s) + 5(-2) This one is most like the first one, except it's a little more spread out. Multiply the term s by the 5 sides in one grouping, and the integer -2 by 5 in the other.

Hope this helps!

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The first steps in writing f(x) = 4x2 + 48x + 10 in vertex form are shown. f(x) = 4(x2 + 12x) + 10 (twelve-halves) squared = 36
brilliants [131]

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f(x)=4(x+6)^2-134

Step-by-step explanation:

We are required to write the functionf(x) = 4x^2 + 48x + 10 in vertex form.

First, bring the constant to the left-hand side.

f(x) -10= 4x^2 + 48x

Factorize the right hand side.

f(x) -10= 4(x^2 + 12x)

Take note of the factored term(4) and write it in the form below.

f(x) -10+4\Box= 4(x^2 + 12x+\Box)

\Box = (\frac{\text{Coefficient of x}}{2} )^2\\\\\text{Coefficient of x}=12\\\\\Box = (\frac{12}{2} )^2 =6^2=36

Substitute 36 for the boxes.

f(x) -10+4\boxed{36}= 4(x^2 + 12x+\boxed{36})

f(x) -10+144= 4(x^2 + 12x+6^2)

f(x) +134= 4(x+6)^2\\f(x)=4(x+6)^2-134

The function written in vertex form is f(x)=4(x+6)^2-134

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