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2 years ago

10

Which equation demonstrates the additive identity property?

1 answer:

Alik [6]2 years ago

7 0

Answer:

The second one

Step-by-step explanation:

Additive identity is adding 0 to a number

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You invest $2000 in a bank account that has 5% annual interest rate, compound ed continously. how much will you have in 5 years?

uysha [10]

Total = $2000(1.05)^5
money = $2552.56

8 0

3 years ago

A tap takes 3 hours to fill a tank. A discharge tap takes 6 hours to empty the same tank. How long will it take to fill the tank

wlad13 [49]

Answer:

2 hours

Step-by-step explanation:

first tap= 3hours

therefore work done is = 1/3

second tap= 6hours

therefore work done = 1/6

so, 1/3+ 1/6

=2/6 + 1/6

=3/6

=1/2

=2/1 hours

=2 hours

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3 years ago

What is the product of 6x-y and 2x-y+2?

Ganezh [65]

Product = multiply

(6x-y)*(2x-y+2) =

12x^2 -8xy + 12x + y^2 - 2y

6 0

3 years ago

Read 2 more answers

Complete parts a through c for the given function.

victus00 [196]

Answer:

a. The critcal points are at

$x=0,-5,3$

b. Then, $x = -5$ is a maximum and $x=3$ is a minimum

c. The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

Step-by-step explanation:

(a)

Remember that you need to find the points where

$f'(x)=0$

Therefore you have to solve this equation.

$20x^4 + 40x^3 - 300x^2 = 0$

From that equation you can factor out $20x^2$ and you would get

$20x^2 ( x^2 +2x - 15) = 0$

And from that you would have $20x^2 = 0$ , so $x = 0$ .

And you would also have $x^2 +2x-15 = 0$ .

You can factor that equation as $x^2 +2x -15 = (x+5)(x-3) = 0$

Therefore $x=-5 , x=3$ .

So the critcal points are at

$x=0,-5,3$

b.

Remember that a function has a maximum at a critical point if the second derivative at that point is negative. Since

$f''(x) = 80x^3 + 120x^2 -600x\\f''(-5) = 80(-5)^3 + 120(-5)^2 -600(-5) = -4000 < 0\\\\f''(3) = 80(3)^3 + 120(3)^2 -600(3) = 1440 > 0 \\$

Then, $x = -5$ is a maximum and $x=3$ is a minimum

c.

The absolute minimum is at $x = 3$ and the absolute maximum is at $x = -5.$

6 0

3 years ago

Read 2 more answers

find the Taylor Series for tan−1(x) based at 0. Give your answer using summation notation and give the largest open interval on

enyata [817]

Let $f(x)=\tan^{-1}x$ . Then $f'(x)=\frac1{1+x^2}$ . Note that $f(0)=0$ .

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n\ge0}x^n$

This means that for $|-x^2|=|x|^2$ , or $-1$ , we have

$\displaystyle f'(x)=\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n\ge0}(-x^2)^n=\sum_{n\ge0}(-1)^nx^{2n}$

Integrate the series to get

$f(x)=f(0)+\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

$\implies\tan^{-1}x=\displaystyle\sum_{n\ge0}\frac{(-1)^n}{2n+1}x^{2n+1}$

6 0

3 years ago