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3 years ago

A cube fish tank has a side length of 12 inches. What is the surface area of the fish tank, including the base and lid?

2 answers:

8 0

72 is correct because a cube has 6 surfaces that are all equal to 12. All you need to do is multiply 6 by 12 and you get your answer.

katrin2010 [14]3 years ago

6 0

72 is the correct answer

You might be interested in

A cylindrical pail that has the base area of 9 pi inches squared and a height of 10 inches. One friend bought a pyramid mold wit

FromTheMoon [43]

Answer:

cylinder — 90π in³
pyramid — 37 1/3 in³
cone — 12.5π in³

Step-by-step explanation:

The volume of a cylinder is given by ...

V = Bh . . . . . where B is the base area and h is the height

The volume of a pyramid or cone is given by ...

V = (1/3)Bh . . . . . where B is the base area and h is the height

The area of a square of side length s is ...

A = s²

The area of a circle of radius r is ...

A = πr²

___

Using these formulas, the volumes of these objects are ...

cylinder: (9π in²)(10 in) = 90π in³

square pyramid: (1/3)(4 in)²(7 in) = 37 1/3 in³

cone: (1/3)(π(2.5 in)²)(6 in) = 12.5π in³ . . . . slightly larger than the pyramid

3 0

4 years ago

Can someone please help me with this! I'll do my best to give a lot of points for it!

Triss [41]

1. By combining like terms
2. Same variable = same term, so you can combine them
3. Because 2y - y = 1y , and 1y is the same as y because a y is one y (obviously)
4. 1/2 + 1/2 = 1. A half an x and a half an x makes one whole x, therefore, x
5. No. You solve them like you would a normal equation with like terms. So it would be 3z^2

Hopefully these are right, I’m a bit rusty!
6. 4x
7. 3y
8. y - 4.5
9. 3.5x + 2
10. 4y + 2
11. 7x
12. 2.2w - 0.5
13. 23b + 6.6
14. 3.75x + 1.5
15. 0.8x + 2.1
Good luck, I hope this helped

8 0

3 years ago

The average annual salary of the employees of a company in the year 2005 was ninety thousand dollars. It increased by the same f

saveliy_v [14]

Given:
Year 0 = 2005 = 90,000
Year 1 = 2006 = 91,200

(91,200 - 90,000)/90,000 = 1,200/90,000 = 0.013

f(x) = 90(1.013)x
f(1) = 90(1.013)1
f(1) = 91.170

The second option, f(x) = 90(1.013)x, is the best representation of the relationship between x and f(x).

8 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=65%20%5Ctimes%2098%20%5Cdiv%2032%20-%2065%20%5Ctimes%207544" id="TexFormula1" title="65 \times

Volgvan

65 * 98 / 32 - 65 * 7544 =
6370 / 32 - 490360 =
199.0625 - 490360 =
-490160.9375

4 0

3 years ago

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