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kenny6666 [7]
2 years ago
14

Please help I give Brainllest, stars, and hearts!

Mathematics
1 answer:
Neporo4naja [7]2 years ago
6 0

Answer is in screenshot. Can I now have Brainliest?

You might be interested in
What is -0.18 as a fraction in simplest form.
Radda [10]
The answer is, 9/50
0.18 = 0.18/1
0.18/1 x 100/100 = 18/100
18/100 divided by 2/2 = 9/50
In conclusion the answer is 9/50.
Hopefully that helped! :)
3 0
3 years ago
-2x+18-1x+4=34 Show your work
Ede4ka [16]

Answer:

x=-4

Step-by-step explanation:

-2x+18-1x+4=34

step 1 combine like terms

-2x+-1x=-3x

18+4=22

-3x+22=34

step 2 subtract each side by 22

-3x=12

step 3 divide each side by -3

x=-4

hope this helps (:

4 0
2 years ago
Read 2 more answers
**Ms. Gayle is driving to work. She travels 10 miles in 10 minutes. How fast is Ms. Gayle traveling?
Dimas [21]

Answer:

60 mph

Step-by-step explanation:

Ms. Gayle drives 10 miles in just 10 minutes. To find out how fast she was going we need to turn the minutes into hours. There are 60 minutes in an hour, so we multiply the miles and minutes by 60.

Ms. Gayle drives 60 miles in an hour.

This means she is traveling at 60 mph.

I hope that this helps! :)

4 0
3 years ago
A line segment has an endpoint at (5,2) and has a negative slope. Which of the following points could be the other endpoint?
slavikrds [6]
<span>C. (4,3)

Draw and plot on a graph and (4,3) could be the other end point, which
x < 5
y > 2</span>
6 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
3 years ago
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