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Anuta_ua [19.1K]
3 years ago
12

Need helpplz and thank you ​

Mathematics
2 answers:
Alex73 [517]3 years ago
6 0

Answer:

Im certain that the answer is D

Step-by-step explanation: HOPE THIS HELPS :)

OlgaM077 [116]3 years ago
5 0

Answer: 2.7 because it’s the introductory

Step-by-step explanation:

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For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusiv
Ugo [173]

Answer:

More than 50

Step-by-step explanation:

To solve, we need to first see that the function is h(n). Picking main points from the question statement:

  • h(n) is the product of all even integers (From 2 to n)
  • p is the smallest factor of h(100)+1
  • h(100)+1 , here n=100

From here, we can write h(100) as:

h(100) = 2 * 4 * 6 * 8 * ...... * 100

h(100) = 2^{50} * (1*2*3*......*50)= 2^{50} * 50!

so,

h(100)+1 =(2^{50} * 50! )+1

Now two numbers,

h(100) and h(100)+1 are consecutive integers and since they are consecutive so they are co-prime. Hence they only have common factor of 1. Example, 13 and 14 have only common factor of 1

As h(100) has all prime numbers from 1 to 50 and according to above statement h(100)+1 won't have any prime factor from 1 to 50, so the smallest prime factor p is greater than 50.

7 0
3 years ago
Greg purchased supplies for a party. He had a coupon good for 30% off. He purchased a helium tank rental with an original price
Kruka [31]
The correct answer is (A) cake
6 0
3 years ago
Read 2 more answers
Write 7,380,000,000 in scientific notation. A. B. C. D.
elena-14-01-66 [18.8K]
7.38 x 10^9 is the correct answer. Hope this helps
7 0
3 years ago
Read 2 more answers
Evaluate using integration by parts ​
PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

I(n)=\displaystyle\int x^ne^x\,\mathrm dx

One round of IBP, setting

u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx

\mathrm dv=e^x\,\mathrm dx\implies v=e^x

gives

\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx

I(n)=x^ne^x-nI(n-1)

This is called a power-reduction formula. We could try solving for I(n) explicitly, but no need. n=5 is small enough to just expand I(5) as much as we need to.

I(5)=x^5e^x-5I(4)

I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)

I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)

I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)

I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))

Finally,

I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C

so we end up with

I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C

I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C

and the antiderivative is

\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C

8 0
3 years ago
(Will give brainliest) Is my answer correct?? If not please correct me
Yuki888 [10]

Hey!

-------------------------------------------------

Answer:

(-6, 2) and (-7, -1) and (0, 0)

-------------------------------------------------

Explanation:

You reflected the triangle. You needed to translate the triangle.

y + 6 is to move up 6 units along the y-axis.

x - 5 is to move left 5 units along the x-axis.

Another thing, the question doesn't state was transformation to use so we are going to assume we are translating.

-------------------------------------------------

Hope This Helped! Good Luck!

5 0
3 years ago
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