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3 years ago

HELP !! giving the brainlest answer to whoever gives the correct answer !! HELPPP PLLSS LAST MINUTEEE

Mathematics

2 answers:

frosja888 [35]3 years ago

4 0

Answer:

Step-by-step explanation:

Vlad1618 [11]3 years ago

4 0

D. I think srry if wrong

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You have 1 pounds of egg whites you need 8oz to make one serving of consomme how many servings can you make

Alexeev081 [22]

Considering that 1 lbs =16 ounces and you need one serving (8oz) then you will make two servings so the answer is two servings hope this helps and have a nice day (:

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Anna71 [15]

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Write an expression that is equivalent to the following: <br><br> x+x+x+x-20

SSSSS [86.1K]

Answer:

4x-20 is the same as x+x+x+x-20

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3 years ago

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What are the solutions of the quadratic function 49x2=9

mylen [45]

Answer:

49x^2 - 9 = 0As there is no x term, we can pretty much guess we have a situation where we factlrise by something known aa difference of two squares, so to factorise it:49 = 7^29 = 3^2x^2 = (x)^2so...(7x - 3)(7x + 3) = 07x - 3 = 0 7x + 3 = 0x = 3/7 x = -3/7

Step-by-step explanation:

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3 years ago

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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