CubeyThePenguin

\(2x^2 - x - 15 = 14\)

\(2x^2 - x - 29 = 0\)

sum = -b/a = -(-1/2) = 1/2

note: sniped

(1)      \((\sqrt{64})^2 = 64\)

(2)        \(\frac{1}{64^{-2}} = 64^2 = 4096\)

(3)        \(64^{1/6} = 2\)

(4)        \(\sqrt{81} \cdot \sqrt{81} = 81\)

The order is (3), (1), (4), (2).

We can use a system of equations. Let t be the cost of the table and b be the cost of the bench.

cost is 685 combined ==> t + b = 685

table costs 35 more than bench ==> t = b + 35

Substitute the second equation into the first.

(b + 35) + b = 685

2b + 35 = 685

b = 325

\((a^3b^5)^4 = a^{12}b^{20}\)

The number of factors is (12+1)(20+1) = 273.

\(\sqrt{\frac{3}{8}} = \frac{\sqrt{3}}{\sqrt{8}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \boxed{\frac{\sqrt{6}}{4}}\)

\(\frac{\sqrt{2.5^2 - 0.7^2}}{2.7 - 2.5} =\frac{\sqrt{(5.76)}}{0.2} = \frac{2.4}{0.2} = \boxed{12}\)

This is a repost, I answered this question already.

\(5(x+2) = x^3 - 2x^2 - 8x\)

\(5(x+2) = x(x-4)(x+2)\)

\(5 = x(x-4)\)                    . (notice that x can't be equal to -2)

\(0 = x^2 - 4x - 5\)

\(\boxed{x = 5, -1}\)

I think the problem implicitly assumes that x is positive, since you're dealing with a physical distance. In that case, the answer would be C.

Here are some reference numbers: \(\sqrt{2021} = 44.955\) and \(\sqrt[3]{2021} = 12.643\).

From 1 to 2021, there are 44 perfect squares and 12 perfect cubes.

From -2021 to -1, there are 12 perfect cubes. (squares can't be negative)

0 is a perfect square and a perfect cube, but you only count it once.