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SVETLANKA909090 [29]

3 years ago

13

P(3,-5)t(x,y)=(x+4,y-3). p'=

1 answer:

ololo11 [35]3 years ago

5 0

Answer:

Step-by-step explanation:

(7,-8)

is the answer

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Sean has to face a big problem but he becomes stronger as a result which saying best summarizes the change in Sean?

kkurt [141]

Answer: D. every cloud has a silver lining

The cloud represents a dark, scary, or sad moment. This is in contrast to the sun which metaphorically represents energy and happiness. The cloud sometimes covers up the sun to cover up those happy moments so to speak.

Despite the cloud being present, there's a silver lining to the cloud which represents some positive aspect a person didn't expect to happen. In this case, the problem allows Sean to get stronger. So this is an opportunity for his growth.

7 0

3 years ago

The triangles are similar.<br> What is the value of x?<br> Enter your answer in the box.<br><br> X=

Aleks04 [339]

Answer:

x=16

Step-by-step explanation:

5times 4=20

3times 4=12

4times 4=16

4 0

3 years ago

A drinking cup in the shape of a cone has a height of 5 in, and a radius of 3.5 in. If the

IrinaK [193]

Geometry helps people understand life u are short and don’t know how life helps u

3 0

3 years ago

A building casts a shadow 168 meters long. At the same time, a pole 5 meters high casts a shadow 20 meters long . What is the he

yawa3891 [41]

Answer:

42

Step-by-step explanation:

The height of the pole is 4 times shorter than the length of the shadow 20÷5=4

The height of the building must be 4 times shorter than its shadow length 168÷4=42

3 0

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

4 years ago