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3 years ago

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Anja simplified the expression StartFraction 10 x Superscript negative 5 Baseline Over Negative 5 x Superscript 10 Baseline EndF

raction to StartFraction 15 Over x Superscript 15 Baseline EndFraction. What mistake did Anja make?

2 answers:

katovenus [111]3 years ago

7 0

Answer:

She subtracted the coefficients instead of dividing them

Step-by-step explanation:

nikdorinn [45]3 years ago

3 0

Answer:

<h2>Anja's mistakes is solving coefficients, she must divide.</h2>

Step-by-step explanation:

The given expression is

$\frac{10x^{-5} }{-5x^{10} }$

Anja simplfied to

$\frac{15}{x^{15} }$

Anja did well powers, however, she did wrong coefficients, instead of dividing them, she subtracted them as she did with powers.

Therefore, Anja's mistakes is solving coefficients, she must divide.

$\frac{10x^{-5} }{-5x^{10} }=-\frac{2}{x^{15} }$

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vesna_86 [32]

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Step-by-step explanation:

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2 years ago

Find the area of the shaded region in square units. Show your reasoning.

motikmotik

Answer:

40 square units

Step-by-step explanation:

First of all, lets say that square has side $l$ , so, the area unit is $l^2$

the diagonal's square is $l\sqrt{2}$

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$A_{square}=side*side=(4\sqrt{2} l)(4\sqrt{2} l)=32l^2$

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3 years ago

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2 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