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2 years ago

What factors do the numbers 20 and 30 have in common?

Mathematics

2 answers:

Amiraneli [1.4K]2 years ago

6 0

Answer:

B. 1, 2, 5, 10

Step-by-step explanation:

Vikki [24]2 years ago

3 0

I think it’s c I hope this works for you

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Rewrite this number in appropriate scientific notation of 4,770,000

Cerrena [4.2K]

The number 4,770,000 in appropriate scientific notation would be 4.770000x10^16

3 0

3 years ago

Plzzzz help asap!!! marking brainiest!!!

Rama09 [41]

Answer:

see below

Step-by-step explanation:

3 ^ ( 6/5)

The base a is 3

The power is 6 which is b

the root is 5 which is c

$\sqrt[5]{3^6}$

3 0

3 years ago

Natalie is making doll customers. Each costume has 4 buttons that coast 3 € each and a zipper that costs 7€. How much does she s

otez555 [7]

24euro on zipper
and 24euro on button

7 0

2 years ago

A rectangle ahs a long side of a length 8.5 cm its perimeter is 24cm work out the length of the shorter side

Phantasy [73]

Answer:

Width (length of shorter side) = 3.5 cm.

Step-by-step explanation:

L = 8.5 cm

perimeter, P = 24 cm

2(L+W) = 24

substitute L = 8.5

2(8.5+W) = 24

8.5 + W = 24/2 = 12

W = 12 - 8.5 = 3.5 cm

5 0

2 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

2 years ago