proyaop

\({1\over a^3+7} = {-a^3\over a^3+7}\)

Multiply both sides by a^3 + 7

1 = -a^3

-1 = a^3

a = -1

3x + 2x = 2(x + 9) - 7 + x

5x = 2x + 18 - 7 + x

2x = 11

x = 11/2

We are looking for x + 1/x = 11/2 + 2/11.

\({11\over2} + {2\over11} = {121\over22} + {4\over22} = {125\over22}\)

k^6 = something that ends in 4.

Notice that something that ends in 4 has to be e^6 where e is even. 6 wont work because 6^n always ends in 6.

2^6 = 64 that works

2^1 = 2

2^2 = 4

2^3 = 8

2^4 = 6

4^6 = 2^12 = 6 so 4 doesnt work

8^6 = 2^18 = 2^[(4*4) + 2] = 4.

So only 2 and 8 work. 12 and 18 also work. 2, 12, 22, 32... 992. 8, 18, 28... 998.

Each of those sets have 100 elements. So there are 200 lucky numbers.

Combine like terms: x^2 + 10x + 25 = 25, subtract 25 from bothsides, x^2 + 10x = 0, you'll see that it can be factored to x(x + 10) = 0, so either of those are equal to 0, if looking for the smaller one, look at x + 10 = 0, x = -10.

-b / 2a = 2

b = -4a

ax^2 - 4ax + c.

a - 4a + c = -1

-3a + c = -1

plug in 2

4a - 8a + c = 4

-4a + c = 4

Subtract equation 2 from equation 1

a = -5

c = -16

b = 20

-5x^2 + 20x - 16

In decimal form, -3/5 = -0.6

Let m = maximum capacity of the aquarium tank, when 2 gallons of water are added to the current m/6 gallons of water, it becomes m/3.

Our equation is \({m\over6 } + 2 = {m\over3}\), multiply all the terms by 6. m + 12 = 2m, so m = 12. 

The total capacity of the aquarium tank is 12 gallons.

1/6 chance to roll a 6 each roll, 3 tries, 3 * 1/6 = 1/2 number of 6s expected to be rolled.

Add both equations together.

-2s + s/2 = 2 + s

s = -4/5

Substitute

3t + 8/5 = 3 - 4/5

3t = 3/5

t = 1/5

The distance from (-4, 6) to (0, 0) is sqrt(4^2 + 6^2) = 2sqrt(13).

The distance from (0, 0) to (4, 3) is 5. So the distance traveled was \(5 + 2\sqrt{13}\) units.