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4 years ago

14. a house-painting company charges 376$ plus 12$ an hour. Another painting company charges 280$ plus 15$ an hour.

1 answer:

5 0

One company charges:
$376 + $12*h

Other company charges:
$280 + $15*h

They will both charge the same when:

$376 + $12*h = $280 + $15*h

96 = 3*h

hours = 32
Cost will be $760

Double Check
$376 + $12 * 32 hours = $760

$280 + $15 * 32 hours = $760

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Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea

Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) $p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

d) $0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

$P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}$

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

$p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}$

And if we simplify we got this:

$p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}$

And if we simplify we got:

$p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}$

And if we find the derivate when t=1 we got this:

$p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114$

And if we replace t=10 we got:

$p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404$

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

$p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And if we set equal to 0 we got:

$0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}$

And then:

$0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)$

$0 =19e^{-\frac{t}{5}} -1$

$ln(\frac{1}{19}) = -\frac{t}{5}$

$t = -5 ln (\frac{1}{19}) =14.722$

7 0

3 years ago

Introduction to Interval Notation What is the domain and range?

Snezhnost [94]

The domain of this function is 3≤x≤5 in interval notation that is [3,5]

The range is -3≤y≤3. In interval notation that is [-3,3]

The domain of this function is -5≤x≤-1 in interval notation that is [-5,-1]

The range is 1≤y≤5. In interval notation that is [1,5]

3 0

4 years ago

WILL MARK BRAINLIEST!

Slav-nsk [51]

Answer:

Second one and the fourth one.

Step-by-step explanation:

Let’s look at this. When we are finding the area of the rectangle, we multiply the length times the width. Let’s look at the first option.

2 times (x + 7)

2x + 14

this one won’t work.

second one:

14(x + 3)

14x + 42

this works.

thrid one.

14(x+5)

14x + 70

nope.

fourth one.

7(2x+6)

14x + 42

works.

fifth one,

2(7x +20)

14x + 40

Does not work.

4 0

3 years ago

Read 2 more answers

The length of the track at talladega superspeedway is 2.66 miles. If the race is 188 laps long , how many kilometers do the driv

Bess [88]

Answer:

Step-by-step explanation:

5 0

3 years ago

A bedroom measures 15 ft by 14 ft with a closet that measures 4 ft by 8 ft. if carpet costs $4.25 per foot, how much will it cos

Anastasy [175]

First of all, I must assume that the correct cost for carpet is $4.25 per square foot.
It could be $4.25 per foot if it came of a standard roll, but then I would need to know the width of the roll, which you haven't given.

Area of the room = (15-ft x 14-ft) = 210 square feet

Area of the closet = (4-ft x 8-ft) = 32 square feet

Total area = (210 + 32) = 242 square feet.

Total cost of carpet = (242 x $4.25) = $1,028.50 plus pad and installation.
=========================================
You know what ? I think you may have mis-copied the price of carpet altogether.
At $4.25 per square foot, that's awfully expensive carpet.
I suspect the actual price is $4.25 per square yard. That's pretty cheap, but
it's a lot closer to reasonable than $4.25 per square foot is ($38.25 per yd² !)

If it's $4.25 per square yard, then we need to know that 1 square yd = 9 square ft.

The total requirement is 242/9 = 26 and 8/9 square yard, so you'll buy 27 sq yds.

$4.25 x 27 = $114.75 plus pad and installation. That sounds cheap but better.

7 0

3 years ago