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sashaice [31]
3 years ago
12

What is the justification for step 2 in the solution process?

Mathematics
1 answer:
igomit [66]3 years ago
3 0

Answer:

y = 11

Step-by-step explanation:

Solve for y

12y - 40 - 4y = 6y - 18

8y -40 = 6y - 18

8y - 6y - 40 = -18

2y - 40 = -18 ( the justification for step 2 is to isolate y on one side of the                   equation in order to solve for y )

2y - 40 + 40 = 40 - 18

2y = 22

2y/2 = 22/2

y = 11

Check :

12(11) - 40 - 4(11) = 6 (11) - 18

132 - 40 - 44 = 66 - 18

48= 48

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To simplify 75/54, we first need to find the greatest common factor. Our GCF is 3, so we will divide both number by 3. 75/3=25, and 54/3=18.
75/54 simplifies into 25/18 or 1 7/18
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Can someone please help me with this question?
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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
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