All categories

2 years ago

Algebra help please !

Mathematics

2 answers:

umka2103 [35]2 years ago

6 0

Answer:

g(f(0)) = 2

Step-by-step explanation:

To find g(f(0)), you must first find f(0).

f(x) = 3x + 1

f(0) = 3(0) + 1 = 0 + 1 = 1

Now you input 1 into function g(x).

g(x) = (4x + 2)/3

g(1) = [4(1) + 2]/3

g(1) = [4 + 2]/3

g(1) = 6/3

g(1) = 2

g(f(0)) = 2

bagirrra123 [75]2 years ago

6 0

For this case we have the following functions:

$f (x) = 3x + 1\\g (x) = \frac {4x + 2} {3}$

We must find g (f (x)). So:

$g (f (x)) = \frac {4 (3x + 1) +2} {3}\\g (f (x)) = \frac {12x + 4 + 2} {3}\\g (f (x)) = \frac {12x + 6} {3}$

Now, with x = 0:

$g (f (0)) = \frac {12 (0) +6} {3} = \frac {6} {3} = 2$

Answer:

Option B

You might be interested in

Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+4y subject to the constraint x2+y2=9, if such values

Vesnalui [34]

The Lagrangian is

$L(x,y,\lambda)=x+4y+\lambda(x^2+y^2-9)$

with critical points where the partial derivatives vanish.

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$

$L_y=4+2\lambda y=0\implies y=-\dfrac2\lambda$

$L_\lambda=x^2+y^2-9=0$

Substitute $x,y$ into the last equation and solve for $\lambda$ :

$\left(-\dfrac1{2\lambda}\right)^2+\left(-\dfrac2\lambda\right)^2=9\implies\lambda=\pm\dfrac{\sqrt{17}}6$

Then we get two critical points,

$(x,y)=\left(-\dfrac3{\sqrt{17}},-\dfrac{12}{\sqrt{17}}\right)\text{ and }(x,y)=\left(\dfrac3{\sqrt{17}},\dfrac{12}{\sqrt{17}}\right)$

We get an absolute maximum of $3\sqrt{17}\approx12.369$ at the second point, and an absolute minimum of $-3\sqrt{17}\approx-12.369$ at the first point.

4 0

2 years ago

Where is the absolute total frequency located in a two-way frequency table

puteri [66]

D. bottom right
Frequency is added here and found out!?

8 0

3 years ago

HELP QUICK! which graph represents the solution set for the inequality PICTURE INCLUDED

grandymaker [24]

The answer is C because it equals 9

5 0

2 years ago

In order to remove water from a flooded basement, two pumps, each rated at 40 gallons per minute, are used. After half an hour,

Scrat [10]

Answer:

3600 gallons of water was removed from the basement.

Step-by-step explanation:

Given:

Rate of each pump = 40 gallons/min

We need to find find the number of gallons of water removed from the basement.

Solution:

Now Given:

After half an hour, the one pump burns out.

Now we know that;

1 hour = 60 mins

$\frac12$ hour = $\frac{60}{2}=30\ mins$

Now for 30 mins both the pumps were working and removing the water.

So we can say that;

Water remove for first 30 mins = $40\times2\times30=2400\ gallons$

Also Given:

the second pump finishes removing the water half an hour later.

So we can say that;

Water removed for next 30 mins = $40\times 30\times1 =1200\ gallons$

Now we can say that;

Total water removed from the basement is equal to sum of Water remove for first 30 mins and Water removed for next 30 mins .

framing in equation form we get;

Total water removed from the basement = $2400+1200=3600\ gallons.$

Hence 3600 gallons of water was removed from the basement.

6 0

3 years ago

Koby can work about 22 math problems in 30 minutes. At this rate how many math problems can Koby work in 3 hours

Murrr4er [49]

132 problems because...

there are six 30 minutes in 3 hours so you would multiply 22 with 6 to get 132
(22*6=132)

4 0

3 years ago

Read 2 more answers