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Anton [14]
3 years ago
14

Scott takes a walk around the block. How far did he walk?

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
7 0
I’m pretty sure it’s 240 because you multiply then add
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DEF rotates 90 degrees clockwise about point A to create D’E’F. Therefore, which equation must be true
Rom4ik [11]

Answer:

m∠EAE ' = m∠FAF '

Step-by-step explanation:

Rotation is a type of transformation referred to a in which the shape of the original object is preserved and the angle of rotation is constant for all the parts of the body.

Thus, when ΔDEF rotates 90° clockwise about point A to create ΔD 'E 'F.

5 0
3 years ago
a wildlife photographer spent 5 minutes taking pictures of a bison at a park. when the bison then decided she didn't want her ph
ZanzabumX [31]

The percent of time the photographer spent negotiating with the bison greater than the time he spent taking pictures is 500%

<h3>Percentage</h3>

  • Time spent taking pictures = 5 minutes
  • Time spent negotiating = 30 minutes

Percentage of time negotiating greater than taking pictures = (difference in time) / time to take pictures × 100

= (30 - 5) / 5 × 100

= 25/5 × 100

= 5 × 100

= 500%

Learn more about percentage:

brainly.com/question/843074

#SPJ1

4 0
2 years ago
PLS HELP ASAP FIRST CORRECT ANSWER GETS BRAINLIEST​
Anna11 [10]
Using Pythagorean, a^2 + b^2 = c^2
16 +x^2 = 36.69
Then subtract the 16 from 36.69, which is 23.69
And finally take the square root of that, giving you 4.87
6 0
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
4 years ago
Multiply.
Inga [223]

Answer:

9 1/11

Step-by-step explanation

I just used Calculator Soup, it helps me with problems that are extremely hard for me to do (look it up)

8 0
3 years ago
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