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3 years ago

PLEASE PLEASE HELP ME PLEASE I AM SO DUM I DONT KNOW THE ANSWER PLEASE JUST HELP ME

Mathematics

1 answer:

pogonyaev3 years ago

3 0

Answer:

Cougar population was 790 at the beginning

830-790=40

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A gourmet pizza café sells three sizes of pizza: small, medium and large. To buy all three sizes, it costs a total of 46.95. If

sveta [45]

Answer:

its gowing to be eleven dollars an 46 sent11.46

6 0

3 years ago

about 47% of the 15152 samples researchers found were bacteria. what percent of the samples were not bacteria?

timofeeve [1]

47% of them were bacteria, which means 53% of them were not bacteria.

47+53=100 or 100-47=53

I hope this helps :)

3 0

3 years ago

Expand and simplify (n-2)(3n+7) + 2n(2n+3)

KonstantinChe [14]

3n^2+7n-6n-14+4n^2+6n= 7n^2+7n-14

Hope that helps and best of wishes!

6 0

3 years ago

The rectangular prism below has a length of 18 inches, a width of 24 inches, and a height of 12 inches.

andrew11 [14]

Answer:

1872

Step-by-step explanation:on edge

3 0

3 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

3 0

3 years ago