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3 years ago

If f(x)=2^x-1 and g(x)= x^2-1, determine the value of (f o g)(3)

Mathematics

1 answer:

zimovet [89]3 years ago

3 0

Answer: 4

Step-by-step explanation:

( f o g ) (3) means f(g(3)) so...

First: plug 3 into g(x) = x^ (2-1)

3^ (2-1) is 3^1 = 3

Next: plug 3 into f(x)= 2^(x-1)

2^(3-1) is 2^(2) > 2×2 = 4

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How many solutions are there to the following equation?

swat32

Answer:

D, infinitely many solutions

Step-by-step explanation:

1. multiply out

5(x-3) = 5x-15

2. combine like terms

5x-15-3x=8x-15-6x = 2x-15=2x-15

they are equal so there are infinite solutions.

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2 years ago

A card is picked at random from the following set: 6, 7, 8, 9. What is P (9)? Write your answer as a percentage.

lozanna [386]

Answer:

25% = (p) 1/4 = picking a number 9.

(p) 1/4 = 0.25 (results getting a number more than 8 or just getting 9

This is add rule from the 4 card set.

1/4 + 0 + 0 + 0 = 1/4

Because each (P) is equal and shows (P); 1/4

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Step-by-step explanation:

0.25 x 100 = 25%

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2 years ago

Read 2 more answers

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madreJ [45]

Answer:

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2 years ago

Here is the sales tax on the same items in City 3.

Dmitry_Shevchenko [17]

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Step-by-step explanation:

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3 years ago

Find the particular solution of the differential equation dy + 7y = 6 satisfying the initial condition y(0) = 0. Answer: y= Your

Aleonysh [2.5K]

Answer:

Step-by-step explanation:

We have the equation $y'+7y=6$ with the initial condition $y(0)=0$ . It is not difficult to notice that this is a linear equation, which has the general expression

$y'+P(x)y=Q(x)$ .

The solution of this equation is expressed by a general formula:

$y(x) = \exp\left(-\int P(x)dx\right)\left(\int Q(x)\exp\left(-\int P(x)dx\right) +C\right)$ .

In the particular case of our equation, we have

$P(x)=7$

$Q(x)=6$ .

Then, we must calculate the integrals

$\int 7dx = 7x$ that implies

$\exp\left(-\int P(x)dx\right) = e^{-7x}$ ,

and

$\int 6e^{7x}dx = \frac{6}{7}e^{7x}$

Then,

$y(x) = e^{-7x}\left(\frac{6}{7}e^{7x} +C\right) = \frac{6}{7} + Ce^{-7x}$ .

In order to obtain the value of the constant we substitute the initial condition

$0=y(0) = \frac{6}{7} + C$ that implies $C=-\frac{6}{7}$

Therefore,

$y(x) = \frac{6}{7}-\frac{6}{7}e^{-7x}$ .

8 0

2 years ago