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bazaltina [42]
3 years ago
14

What is the scale factor of the dilation?

Mathematics
1 answer:
vodomira [7]3 years ago
6 0
I think it's A when shown work it equals to 2 not -1/2
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All of the equations to –32 = k + –19
soldi70 [24.7K]

Answer:

k =  - 13

Step-by-step explanation:

- 32 = k - 19 \\  -32 + 19 =  k \\  - 13 = k

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A wildlife service wanted to estimate the number of alligators in a river so they tagged 90 alligators and released them back in
lilavasa [31]
You can solve this with a proportion. If you know that 23 tagged alligators were in a 200 alligator sample, there are 90 tagged alligators in the entire river. Written otherwise:

23/200 = 90/x    

and solve for x:

x = 782.608...

So the "best estimate" would be around 783 alligators. Hope that helps! If you liked this answer please rate it as brainliest!! thank you!!!
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4 years ago
Find the lengths of the missing sides. If the length is irrational, round to the nearest tenth. ​
grigory [225]

Answer: 25

Step-by-step explanation:

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PLZ, help me with this question.
GenaCL600 [577]

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Express the integral as a limit of Riemann sums. Do not evaluate the limit. (Use the right endpoints of each subinterval as your
Veronika [31]

The expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Given an integral \int\limits^5_b {1} \, x/(2+x^{3}) dx.

We are required to express the integral as a limit of Riemann sums.

An integral basically assigns numbers to functions in a way that describes displacement, area, volume, and other concepts that arise by combining infinite data.

A Riemann sum is basically a certain kind of approximation of an integral by a finite sum.

Using Riemann sums, we have :

\int\limits^b_a {f(x)} \, dx=\lim_{n \to \infty}∑f(a+iΔx)Δx ,here Δx=(b-a)/n

\int\limits^5_1 {x/(2+x^{3}) } \, dx=f(x)=x/2+x^{3}

⇒Δx=(5-1)/n=4/n

f(a+iΔx)=f(1+4i/n)

f(1+4i/n)=[n^{2}(n+4i)]/2n^{3}+(n+4i)^{3}

\lim_{n \to \infty}∑f(a+iΔx)Δx=

\lim_{n \to \infty}∑n^{2}(n+4i)/2n^{3}+(n+4i)^{3}4/n

=4\lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3}

Hence the expression of integral as a limit of Riemann sums of given integral \int\limits^5_b {1} \, x/(2+x^{3}) dxis 4 \lim_{n \to \infty}∑n(n+4i)/2n^{3}+(n+4i)^{3} from i=1 to i=n.

Learn more about integral at brainly.com/question/27419605

#SPJ4

5 0
2 years ago
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