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3 years ago

{(1, 4), (2, 4), (3, 2)} is a function?

Mathematics

1 answer:

kompoz [17]3 years ago

4 0

1. yes, it is a function

2. { 1,2,3} = domain

3. { 2,4 } = range

4. 5x...when x = 4......5(4) = 20

5. True

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Which theorem would you use to prove ABE ~ DCE?

Pachacha [2.7K]

Answer: Choice C) AA similarity

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We are given that angle ABE = angle ECD since they are both right angles (square marker)

Angle AEB = angle CED as they are vertical angles

we have two pairs of congruent angles, leading us to be able to use the AA (angle angle) similarity theorem

3 0

3 years ago

3/4+7/12<br>adding and subtracting fractions with value

lesantik [10]

Answer:It would be 4/3 or 1 1/3. You can put 3/4 to 9/12 and then add.

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

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$

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3 years ago

1 - (9 – 4) : 5 =<br> Please someone help

Norma-Jean [14]

Answer:

-4 = 5

Step-by-step explanation:

9-4=5

1-5=-4

-4:5

3 0

4 years ago

Find the linearization L(x) of the function at a. f(x) = x^4 + 2x^2, a = 1

gregori [183]

The equation of the tangent line at x=1 can be written in point-slope form as

... L(x) = f'(1)(x -1) +f(1)

The derivative is ...

... f'(x) = 4x^3 +4x

so the slope of the tangent line is f'(1) = 4+4 = 8.

The value of the function at x=1 is

... f(1) = 1^4 +2·1^2 = 3

So, your linearization is ...

... L(x) = 8(x -1) +3

... L(x) = 8x -5

8 0

3 years ago