NotThatSmart

We basically have all the information needed to solve this problem. 

First off, let's note that 

\((x+y)^3 = x^{3}+3x^{2}y+3xy^{2}+y^{3}\)

Simplifying this, we get

\(8=5+6xy\\ xy=1/2\)

Now, let's note that \((x+y)^2 = x^2+2xy+y^2\)

This contains x^2+y^2, so let's isolate that. We get

\(x^2+y^2 = (x+y)^2 - 2xy\)

We already have all the values shown for us to find x^2+y^2. Plugging in 2 and 1/2, we get

\(x^2 + y^2 = 4 - 1 = 3\)

So 3 is our final answer. 

Thanks! :)

There actually is no minmum for y. 

The range for y is \(f\left(x\right)\le \frac{1}{4}\) as there is no limits to how small it can be. 

There is a maximum of 1/4, as the max point is

\(\left(1,\:\frac{1}{4}\right)\)

Thanks! :)

We can split this equation into 2 equations. Since if either them equal 0 the equation will be true, we just set both quadratics to equal 0. 

We have

\(2x^2+4x+2=0\quad \mathrm{or}\quad \:3y^2-6y+3=0\)

Solving the first equation, we have

\(2(x^2+2x+1)=0\\ 2(x+1)^2=0\\ x=-1\)

Solving the second equation, we have

\(3(y^2+-2y+1)=0\\ 3(y-1)^2=0\\ y=1\)

So there are an infinite amount of ordered pairs for x and y as long as either \(x=-1\) or \(y=1\)

So infinite is our answer...

Thanks! :)

In that case, we can do the same with 7 and 8. 

We have 

\(\binom{1 + 4 - 1}{4 - 1} = \binom{4}{3} = 4\)

and

\(\binom{0 + 4- 1}{4 - 1} = \binom{3}{3} = 1\)

So there are 10+4+1 = 15. 

So our answer is 15/495. 

Thanks! :)

We want bc to be the biggest, ac to be the second biggest, and ab to be the smallest. 

Obviously, c has to be the biggest number and b has to be the second biggest number. 

Thus, c is 3, b is 2, and a is 1. 

We have

\((2*1) + 2(1)(3) + 3(3)(2)\\ 2+6+18\\ 26\)

So our answer is 26. 

Thanks! :)

Let's convert the each number in the fraction to base 25. 

One always stays the same, so converting 288 to base 25, we have

\((BD)_{25} = (11 × 25^1) + (13 × 25^0) = (288)_{10}\)

This means that

\(\frac{1}{288}_{10} = \frac{1}{BD}_{25}\)

Now, we do base division. We get that

\(\frac{1}{BD}_{25} = 0.\overline{02468ACEGIKN}\)

As you can see, 1/288 in base 25 is repeating. 

So repeating is our answer. 

Thanks! :)

Let's split this into two inequalities and solve both to find what x is. 

We have

\(64 \leq n^2\)

and 

\(n^2 \leq 100+20n\)

Let's solve the first equation. Square rooting both sides, we get

\(n \geq 8\) or \(n \leq -8\)

Solving the second equation, we get that

\(n^2-20n-100 \leq 0\\ 10\left(1-\sqrt{2}\right) \leq n\leq10\left(1+\sqrt{2}\right)\)

Merging overlapping intervals, we have

\(8\le \:n\le \:10\left(1+\sqrt{2}\right)\)

This rounds to 

\(8\le \:n\le \:24.14\)

So our answer is 17. 

Thanks! :)

First, let's combine all like terms in the equation. We have

\(4x^{4}{-10x^{2}}\)

Factoring this equation, we get that

\(2x^{2}\left(2x^{2}-5\right)\)

Now, we can graph this. We have that

So the minimum value is -6.25. 

So our final answer is -6.25 

Thanks! :)

We can't go ahead and directly solve for x, so let's take a different approach. 

Let's first set \(u=\sqrt[3]{x+14}\). We can solve for u, then plug it in to solve for x. 

For this value of u, we get \(x=u^3-14\)

Subsituting out all x, we get that 

\(u-\sqrt[3]{\left(u^3-14\right)-14}=\sqrt[3]{4}\)

Removing the roots, we have 

\(-u^3+28=4-3\cdot \:4^{\frac{2}{3}}u+3\sqrt[3]{4}u^2-u^3\)

Combining all like terms, we ge tthat

\(3\sqrt[3]{4}u^2-3\cdot \:4^{\frac{2}{3}}u-24=0\)

Using the quadratic equation, we have

\(u=\frac{4^{\frac{2}{3}}+2\left(8\sqrt[3]{4}+4^{\frac{1}{3}}\right)^{\frac{1}{2}}}{2\sqrt[3]{4}},\:u=\frac{4^{\frac{2}{3}}-2\left(8\sqrt[3]{4}+4^{\frac{1}{3}}\right)^{\frac{1}{2}}}{2\sqrt[3]{4}}\)

Now subbing this value of u and finding x, we have

\(\sqrt[3]{x+14}=\frac{4^{\frac{2}{3}}+6\sqrt[3]{2}}{2\sqrt[3]{4}} \\ x+14=\frac{55}{2}+\frac{9}{2}\\ x=18\)

and 

\(\sqrt[3]{x+14}=\frac{4^{\frac{2}{3}}-6\sqrt[3]{2}}{2\sqrt[3]{4}}\\ x+14=\frac{1}{2}-\frac{9}{2}\\ x=-18\)

So our answers are 18 and -18. 

I completed this in a really complicated manner. I'm interested to see if there was a more efficient way to complete this.

Thanks! :)