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2 years ago

14

8 The perimeter P of a regular pentagon is 90 cm. What is the length x of each side?

2 answers:

svetoff [14.1K]2 years ago

8 0

Answer:

X = 18

Step-by-step explanation:

This is a regular pentagon, meaning that each side is the same. Now, since we know the perimeter, we can divide that by 5 to find each side's length (x). 90/5 = 18, so the length x of each side is 18.

Artyom0805 [142]2 years ago

5 0

Answer:

18cm

Step-by-step explanation:

You would do 90/5 which is 18. Each side of the pentagon is 18cm.

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For the graph, what is a reasonable constraint so that the function is at least 200?

Anastasy [175]

Answer:

$0\leq x\leq 15$

Step-by-step explanation:

I think your question missed key information, allow me to add in and hope it will fit the orginal one

<em>For the graph below, what should the domain be so that the function is at least 200? graph of y equals minus 2 times the square of x plus 30 times x plus 200 </em>

My answer:

Given the above information, we have:

$y=-2x^2+30x+200$

To make the function is at least 200, it means that:

$y=-2x^2+30x+200$ ≥ 200

<=> $-2x^2+30x$ ≥ 0

<=> x(-2x+30) ≥ 0

This is the product of two numbers hence would be positive only if either both are positive or both are negative

Case I: Both positive

x ≥ 0 and (-2x+30) ≥ 0

<=> 0 ≤ x ≤ 15

Case II: Both negative

Then we get

$x\leq 0 and -2x+30\leq 0\\\\x\leq 0 and x\geq 15$

This is inconsistent as a value cannot be less than 0 and greater than 15

=> our correct answer is $0\leq x\leq 15$

Hope it will find you well.

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Step-by-step explanation:

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pantera1 [17]

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$F(x)=\displaystyle\sum_{k\ge0}k^3x^k$

(if we include $k=0$ for a moment)

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k$

Take the derivative to get

$\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k$

$\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k$

Take the derivative again:

$\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k$

$\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k$

Take the derivative one more time:

$\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k$

$\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k$

so we have

$\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}$

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