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4 years ago

I’ll mark brainlest:( i need help!

Mathematics

1 answer:

enyata [817]4 years ago

7 0

Answer: Choice A) $3 \sqrt{26}$

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Work Shown:

$\text{The two points are } (x_1,y_1) = (6,7) \text{ and } (x_2,y_2) = (3,-8)\\\\d = \text{Distance between } (x_1,y_1) \text{ and } (x_2,y_2)\\\\d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} = \text{Distance Formula}\\\\d = \sqrt{(6-3)^2+(7-(-8))^2}\\\\d = \sqrt{(6-3)^2+(7+8)^2}\\\\$

$d = \sqrt{(3)^2+(15)^2}\\\\d = \sqrt{9+225}\\\\d = \sqrt{234}\\\\d = \sqrt{9*26}\\\\d = \sqrt{9}*\sqrt{26}\\\\d = 3\sqrt{26}\\\\$

When going from 234 to 9*26, the idea here is to factor where one factor is the largest perfect square possible. That way we can use the rule $\sqrt{x*y} = \sqrt{x}*\sqrt{y}$ to break up the root and simplify.

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Find a compact form for generating functions of the sequence 1, 8,27,... , k^3

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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3 years ago

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Step-by-step explanation:

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3 years ago

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For this case, we have that:

By definition, a rectangle is a parallelogram. So, we discard the first option.

On the other hand, we have that all rectangles can not be squares, since a square has all its sides equal, while the rectangle, poir definition, no.So, we discard the second option.

All quadrilaterals are not rectangles because there are other figures with four sides that are not necessarily rectangular. We discard the third option.

Finally, we have that all squares are rectangles. This statement is true since squares can be rectangles with pairs of equal sides.

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