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3 years ago

What is your grade? I want to know if mine is average.

Mathematics

2 answers:

Advocard [28]3 years ago

6 0

— 10th grade! Hbu? —

Stells [14]3 years ago

4 0

10th grade what bout you

You might be interested in

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

On Monday night Eli ate 1/2 of a pizza. On Tuesday night, Eli ate 1/4 of a pizza. How much pizza did he eat altogether

-BARSIC- [3]

Answer:

Eli ate 3/4 of the pizza

Step-by-step explanation:

1/2 = 2/4

2/4 + 1/4 = 3/4

5 0

3 years ago

Read 2 more answers

Simplify (-2)^-3<br> a. -8<br> b. -1/8<br> c. 1/8<br> d. 8

dusya [7]

Answer:

Its D because 2×2×2 equals 8 and they are both minus and knowing - times - makes + so it becomes 8

7 0

3 years ago

Read 2 more answers

Allegra has a cell-phone plan that charges $65 per month and $0.10 for every minute that she uses the phone beyond what her plan

Bond [772]

It seems confusing, but it's just breaking down the steps for you. It's done the same as any other word problem.
a. The unknown is how many minutes she used beyond the plan. So, we'll use "m" for the variable, to represent minutes.
b. 73.40 = 65 + 0.10m
   The 73.40 is the total, which is why it's by itself. 65 represents the price per month for the plan, and 0.10m is the price per extra minute multiplied by the unknown amount of minutes used.
c. 73.40 = 65 + 0.10m
   8.40 = 0.10m
   84 = m
d. The number of minutes that Allegra went over the time that the plan allows is found by solving the equation we just wrote and solved. $0.10 is paid for every minute past the plan's allowance, meaning that "m" in the equation, when solved, shows us exactly how many minutes over Allegra went.

8 0

3 years ago

The base of a side of a ramp is 3 meters and the height os 0.6 Meters what us the area

Kobotan [32]

The answer would be

area = base x height divided by 2

3*0.6 = 1.8/2=0.9 square meters

5 0

4 years ago