NotThatSmart

There are actually many different numbers that satisfy this equation. 

First off, we have

\(244224 \mod 64 = 0 \)

So 244224 works. 

Also, we have

\(444224 \mod 64 = 0\)

So 444224 also works. 

So our answers are \(444224, 244224\)

Thanks! :)

Let's first multiply the two functions to make our lives really easy. We have

\(( -x^2+ 8x - 5)(x^3 - 11x^2 + 2x)\\ -\left(x^{5}-11x^{4}+2x^{3}\right)+8x\left(x^{3}-11x^{2}+2x\right)-5\left(x^{3}-11x^{2}+2x\right)\\-x^{5}+19x^{4}-95x^{3}+71x^{2}-10x\)

The leading term is \(-x^5\)

The leading coefficient is \(-1\)

The degree of the polynomial is the highest power appearing in the polynomial, which in this case, is 5. 

The constant term is the one with no x value. Since there is no constant, the constant is just 0. 

The coefficient of x^2 is \(71\)

Thanks! :)

Let's note something really important before we begin. 

To solve this problem, we must first do something quite important.

It's crucial that we split this inequality into two seperate ones before combining them back again. Thus, we have

\(x^4+4x^2>-4\\ x^4+4x^2<21\)

Now, note something about the first inequality. 

We have the terms x^4 and x^2 which is guarenteed to be positive. Since we are adding them, all terms of x work, so we don't worry about anything. 

Next, we have

\(x^4+4x^2<21\)

Subtracting both sides by 21 and rewriting in standard form, we have the inequality

\(x^4+4x^2-21<0\)

Factoring out +-\sqrt3, we have

\(\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)\left(x^2+7\right)<0\)

Now, let's note the intervals. If x is greater than square root 3, then we have a positive number. If x is less than square root 3, we also have a positive number. 

If x is either, then we have 0. Thus, the interval is

\(-\sqrt{3} <\sqrt{3}\)

In interval notation, we have

\((-\sqrt{3},\sqrt{3})\)

So our answer is \((-\sqrt{3},\sqrt{3})\)

Thanks! :)

Modulo is essentially the remainder we get when we divide two numbers.

For example, we have

\(10 \equiv 1 (\mod 3)\)

In this case, 1 is the remainder, 3 is the number we divide into 10. 

Thus, we say that 10 is congruent to 1 modulo 3. 

To solve the problem, we have the equation

\(9n+1 \equiv 1 (\mod 9)\)

We simply can write the equation

\(9n+1 \leq 1000\)

Now, we simplfy solve for n. Setting the two equations togther, we have

\(9n = 999\\ n=111\)

Thus, our answer is 111. 

Thanks! :)

Thanks, CPHill. 

I just realized you could have just done \(100-25 = 75\%\) to find the total amount of orange and dark. 

Then, we multiply the reciprocal of that by \(78\)

It gets the same answer of \(104\) since it's essentially the same steps. We still have

\(x = 78 * 4/3\\ x = 26*4\\ x = 104\)

It does save a few steps and is more efficient, but hey, same answer...

Thanks! :)

Let's set up some variables to solve this problem. 

Let's first set that \(P = (x,y)\)

Now, from the problem, we can write the complicated equation of

\(PA^2 + PB^2 + PC^2 = (x-7)^2 + (y + 11)^2 + (x-10)^2 + (y-13)^2 + (x-18)^2 + ( y + 22)^2 \)

Now, we have to simplify this gigantic equation. By expanding and combining lots of like terms, we get that

\(3 x^2 - 70 x + 3 y^2 + 40 y + 1247 \\ 3 [ x^2 - 70/3 x + y^2 + 40/3 y ] + 1247\)

Now, we simplfy complete the square for x and y to achieve our x and y values for point Q, which will be crucial. 

Completing the square, we have

\(3[ x^2 - 70/3 x + 4900/36 + y^2 + 40/3y + 1600/36 ] + 1247 - 4900/12 -1600/12 \\ 3 [ (x - 70/6]^2 + (y + 40/6)^2 ] + 2116 / 3\)

Now, from this, we can tell what Q has for their x and y value. 

We have that Q = (70/6, -20/3)

Thus, we know that

\(k = 2116 / 3 \)

So 2116/3 is our final answer. 

Thanks! :)

Very smart way to do it!

I totally forgot about factoring a^3 + b^3 as \((a+b)(a^2-ab+b^2)\)

It makes a lot of sense and is probably the most efficient way to complete this problem. 

Subsituting in \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3} = a^3 + 3ab(a+b)+b^3\) work, I geuss, but good work!

Both solutions do eventually reach the desired goal of 24, so that's awesome. :)

Good job to you too, @aboslutelydestroying!

We don't actually have to find what a and b is. 

First, let's note that we have \((a+b)^2 = a^2+2ab+b^2\)

Squaring both sides of the first equation, we find that

\(a^2+2ab+b^2 = 16\)

Subtracting the second equation from this equation we just computed, we have

\(3ab = 10\\ ab = 10/3\)

Now, let's note something else real quick. 

We have that \((a+b)^3=a^{3}+3a^{2}b+3ab^{2}+b^{3} = a^3 + 3ab(a+b)+b^3\)

Since we also know that \((a+b)^3 = 4^3=64\), we have the equation \( a^3 + 3ab(a+b)+b^3=64\)

Wait! we already have all the numbers needed to isolate a^3 and b^3. We just found what ab was and a+b is given, so we have

\(a^3+b^3=64-3(10/3)(4) \\a^3+b^3 = 64-40\\ a^3+b^3=24\)

So 24 is our final answer. 

Thanks! :)

I'm not the best at these questions, but heres my understanding. We have

\(f_Q(x)=\sqrt{5^2-x^2}\\ f_R(x)=\sqrt{8^2-(x-11)^2}\\ 25-x_P^2=64-x_P^2+22x_P-121\\ x_P=\dfrac{121+25-64}{22}\\ x_P=3.\overline{72}\\ y_P=3.333\)

We are using coordinate geometry to solve this problem. Now, since we know these two values, we find that

\(P(3.\overline{72},\ 3.333)\\ M(5.5,\ 0)\\\)

Now, we can use the distance formula to find PM. We have

\( \overline{PM}=\sqrt{(x_M-x_P)^2+y_P^2}=\sqrt{(5.5-3.727)^2+3.333^2}\\\overline{PM}=3.775\)

I don't think this was the clearest explenation, but that's all I could understand. 

So 3.775 is our answer. 

Thanks! :)

We can use a handy equation and the fact that it's a right triangle to solve this problem. 

First, let's note some equations. Since AM and BN is the median, let's set a and b to be the two legs of the triangle. we have the equation

\(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)

\(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)

Now, these two equations are quite important, since we want to know what a^2+b^2 is. Plugging in 1 and 1, and isolating all terms except for a and b, we find that

\(2^2 = 4a^2+b^2\\ 2^2 = 4b^2 + a^2\)

Now, adding the two equations, we have the formula

\(5a^2+5b^2 = 8\\ a^2+b^2 = 8/5\)

Thus, the hypotenuse, which is equal to the square root of a^2 + b^2 is just \(\sqrt{8/5}\)

So our answer is just \(\sqrt{8/5}\)

Thanks! :)