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3 years ago

Can yall please help me, try to explain in steps, tyy

Mathematics

1 answer:

almond37 [142]3 years ago

8 0

Answer:

b. would be the answer bc it goes in a line if you were to connect the dots

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What are the domain and range of the exponential function below? <img src="https://tex.z-dn.net/?f=F%28x%29%20%3D%205%5

faust18 [17]

Answer:

Step-by-step explanation:

The domain is the set of x values for which the function is defined.

The range is the set of y values for which the function is defined.

Attached is the graph of the exponential function.

It is the basic graph of exponential function of y = 5^x which is shifted 6 units above (because of +6 at the end).

Looking at the graph, the domain is the set of all x values.

The range is anything above 6.

Correct answer is B.

3 0

3 years ago

Need help asap please

Rzqust [24]

Answer:

Option 3

Step-by-step explanation:

The number on the inside is the dividend and the number on the outside is the divisor. So we just have to place the dividend on the top and the divisor on the bottom.

7 0

3 years ago

How long will it take for an investment to triple, if interest is compounded continuously at 4%?

Margaret [11]

3p=pe^tr,,3=e^0.04t ,,then use log to find t

3 0

3 years ago

Let f be defined as shown. What is f^-1(1)?

FinnZ [79.3K]

Answer:

From the above image when the input of f(x) is 9 the output is 1

So f^-1(1) is just the reverse when the input is 1 the output is 9

So f^-1(1) = 9

Hope this helps

6 0

3 years ago

A tank initially has 300 gallons of a solution that contains 50 lb. of dissolved salt. A brine solution with a concentration of

belka [17]

Let s(t) be the amount of salt in the tank at time t. Then s(0) = 50 lb.

Salt flows into the tank at a rate of

(2 gal/min) (6 lb/gal) = 12 lb/min

and flows out at a rate of

(2 gal/min) (s(t)/300 lb/gal) = s(t)/150 lb/min

so that the net rate at which salt is exchanged through the tank is

ds(t)/dt = 12 - s(t)/150 … (lb/min)

Solve for s(t). The DE is separable, so we have

ds/dt = 12 - s/150

150 ds/dt = 1800 - s

150/(1800 - s) ds = dt

Integrate both sides to get

-150 ln|1800 - s| = t + C

Solve for s :

ln|1800 - s| = -t/150 + C

1800 - s = exp(-t/150 + C )

1800 - s = C exp(-t/150)

s = 1800 - C exp(-t/150)

Now given that s(0) = 50, we solve for C :

50 = 1800 - C exp(-0/150)

50 = 1800 - C

C = 1750

Then the amount of salt in the tank at any time t ≥ 0 is

s(t) = 1800 - 1750 exp(-t/150)

To find the time it takes for the tank to hold 100 lb of salt, solve for t in

100 = 1800 - 1750 exp(-t/150)

1700 = 1750 exp(-t/150)

34/35 = exp(-t/150)

ln(34/35) = -t/150

t = -150 ln(34/35) ≈ 4.348

So it would take approximately 4.348 minutes.

By the way, we didn't have to solve for s(t), we could have instead stopped with

-150 ln|1800 - s| = t + C

Solve for C - this C is not the same as the one we found using the other method. s(0) = 50, so

-150 ln|1800 - 50| = 0 + C

C = -150 ln|1750|

==> t = 150 ln(1750) - 150 ln|1800 - s|

Then s(t) = 100 lb when

t = 150 ln(1750) - 150 ln(1700)

t = 150 ln(1750/1700)

t = 150 ln(35/34)

t ≈ 4.348

5 0

3 years ago