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2 years ago

HELP APPRECIATED WILL GIVE BRAINLIEST:)

Mathematics

2 answers:

klemol [59]2 years ago

7 0

Answer:13 1/8

(Decimal: 13.125)

Step-by-step explanation:

vodomira [7]2 years ago

4 0

Answer:18 5/8

Step-by-step explanation:

Rewriting our equation with parts separated

Solving the whole number parts

Solving the fraction parts

Find the LCD of 1/2 and 1/8 and rewrite to solve with the equivalent fractions.

LCD = 8

Combining the whole and fraction parts

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horrorfan [7]

Answer:

y = 2/3x+1/3

Step-by-step explanation:

2x−3y+1 = 0

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2x−3y+1 +3y= 0+3y

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2 years ago

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Step-by-step explanation:

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2 years ago

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3 years ago

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

pantera1 [17]

This sequence has generating function

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

(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:

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

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so we have

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

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