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stealth61 [152]

2 years ago

6

The ordered pair ( x + 4, y + 6) was used to translate an object. To find the pre-image of the object, you would need to move __

___.
four units right and six units up
four units left and six units down
four units left and six units up
six units up and four units down

1 answer:

Harman [31]2 years ago

5 0

B four units left and six units down

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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

2 years ago

you take a group of 20 people going to the movies. kids’ tickets cost $4 each, and adults tickets $6 each. your total cost is $9

brilliants [131]

The number of kids in your group is 12.

Step-by-step explanation:

Given,

Number of people = 20

Cost of each kid ticket = $4

Cost of each adult ticket = $6

Total spent = $96

Let,

x be the number of kids ticket

y be the number of adult ticket

According to given statement;

x+y=20 Eqn 1

4x+6y=96 Eqn 2

Multiplying Eqn 1 by 4

$4(x+y=20)\\4x+4y=80\ \ \ Eqn\ 3$

Subtracting Eqn 3 from Eqn 2

$(4x+6y)-(4x+4y)=96-80\\4x+6y-4x-4y=16\\2y=16$

Dividing both sides by 2

$\frac{2y}{2}=\frac{16}{2}\\y=8$

Putting y=8 in Eqn 1

$x+8=20\\x=20-8\\x=12$

The number of kids in your group is 12.

Keywords: linear equation, elimination method

Learn more about elimination method at:

#LearnwithBrainly

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