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2 years ago

Write 1 2/3 as the sum of a whole number and two fractions that have the same denominator.

Mathematics

1 answer:

larisa [96]2 years ago

7 0

1+1/3+1/3 because 1/3 and 1/3 have the same denominator.

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® Incorrect<br> Which would show that FGHI is a parallelogram? Select all that apply.

Tatiana [17]

Answer:

FGHI is a Parallelogram

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2 years ago

Felicia wants to buy a new soccer ball. it is on sale for $12.60 She has one $10 bill,two $5 bills,three$1 bills,6 quarters,and

gladu [14]

One $10 bill, Two $1 bills, Two Quarters, Two Nickels

Two $5 bills, Two $1 bills, Two Quarters, Two Nickles

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3 years ago

Stephen rode his bike 23.4865 miles On monday and 38.243 miles on Tuesday. How many miles did he ride in all

professor190 [17]

Answer:

61.7295 (61.7) miles

If I'm reading this correctly, this is just a simple addition equation. You'd add up 23.4865 and 38.243 to get 61.7295 miles, or 61.7 miles.

Hope I helped! ☺

6 0

3 years ago

Can u pls help i have a few mins left

BartSMP [9]

Answer:

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

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