All categories

2 years ago

What is 288889 divide by 676767 please help

Mathematics

2 answers:

Svetach [21]2 years ago

8 0

The correct answer is 0.426866262687158

professor190 [17]2 years ago

7 0

Answer:

0.42686626268 yeah that it

You might be interested in

Alya loves to read. She read 90 pages in half an hour. How many pages does she read per minute?

IgorC [24]

Answer: 3

90 pages in 30 minutes so divide 90 by 30 and 90 divided 30 = 3

Hope this helps!

6 0

3 years ago

Read 2 more answers

2(x – 2) + 6 = 0 <br> can anyone tell me the answer with steps please

Andrews [41]

Answer:

x = -1

Step-by-step explanation:

2(x – 2) + 6 = 0

~Distribute left side

2x - 4 + 6 = 0

~Combine like terms

2x + 2 = 0

~Subtract 2 to both sides

2x = -2

~Divide 2 to both sides

x = -1

Best of Luck!

5 0

3 years ago

A rectangular garden is fenced on all sides with 278 feet of fencing. The garden is 9 feet longer than it is

JulijaS [17]

It’s 301 bc u add them together

6 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

2/8 + 6/12 what is that eqaul to

Papessa [141]

$\frac{2}{8} + \frac{6}{12} \\ \\ \frac{1}{4} + \frac{1}{2} \\ \\ LCD = 4 \\ \\ \frac{1}{4} + \frac{2}{4} \\ \\ \frac{1+2}{4} \\ \\ \frac{3}{4} \\ \\ Answer: \fbox {3/4} \ or \ \fbox {0.75}$

8 0

3 years ago

Read 2 more answers