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3 years ago

15

Simplify the following expression using only positive exponents. Help please!? It has to be only positive exponents the -9/7t an

swer will not work

1 answer:

wlad13 [49]3 years ago

8 0

(-9t^25) / (7t^26) = -9 / (7t)

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Please i need help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

tekilochka [14]

Answer:

limeted government or rule of law i think sorta mayby kindaa

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2 years ago

(3x - 1)+(x ^ 2 + 5x - 4)

RideAnS [48]

Answer: if it is simply then the answer is 3x^3+14x^2-17x+14

Step-by-step explanation:

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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:

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

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

so we have

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

5 0

3 years ago

What is the simplified expression for 2 (x minus 5)? 2 x Negative 10 x 2 x minus 10 2 x + 5

Kay [80]

Answer:

2x-10

Step-by-step explanation:

Multiply BOTH x and -5 by 2

7 0

3 years ago

Read 2 more answers

Help Asap Immediately!!

Svetlanka [38]

Answer:

its 432

Step-by-step explanation:

You multiply the base times the side length then divide by 2. hope this helps

3 0

2 years ago