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Also, please consider looking for this problem rather than making a new webcalc question because perhaps this question has been posted already and answered already.

These are a system of equations. Add the first two equations, and you get 2a^3 (the 3ab^2 cancel each other out) = 1352. a^3 = 676. 

a = cube root of 676.

Plugging back in, we have 676 + 3cbrt(676)b^2 = 679

cbrt(676)b^2 = 1

1/cbrt(676)

b = sqrt(cbrt(676^2)/676) = cbrt(676)/26

a - b = cbrt(676) - cbrt(676) = 25cbrt(676)/26

I would make sure you got your measurements correct. Triangle BDI is impossible by the triangle inequality. Maybe a typo?

We can use proof by contradiction here

We can first assume first that they are the same function, which means that for every single x value there is the same y value. It is easy to see that is not the case here, for example if f(5) is -5, while h(5) is -20. Because of this, that means the statement we assumed at the beginning must be false, so they are not the same function.

TooEasy, do you like to \(\text{su} \text{ck} \) \(\text{co} \text{ck}\)?

Here's a tip, go to this link:
 

https://web2.0calc.com/questions/algebra_55659

Beautiful answer by @CPhill

Im not sure either

That wasnt right either 

Because QN is a perpendicular midpoint, it is basically an altitude

We also know the base is 14, so the area of the triangle is \(12 \times 14 \div 2 = \color{brown}\boxed{84}\)

I like your solution, for a second I thought you got it wrong nvm

I agree with your solution, instead I did this another way

I started with considering all the possible numbers this could be equal to because of the inequality sign, 0 1 or 2

For zero there can only be 1 possibility, all zeros

For x_1+x_2...x_10 equal to 1, there are 10 possibilities, one of them is one, the others are zero 

For x_1+x_2...x+10 to be equal to 2, we use stars and bars (we could have done for the first two cases, but I'm too lazy ), so it would be (10+2-1) choose (10-1) which is 11 choose 9 which is 55

Adding all of these together we get 55+10+1 which is 66

The answer is 10.

The probability is 17/200.

Strange... that's what I got as well, and Mathpapa verifies that

This is a pretty classic problem.

First, we can set variables for how many stickers each of his friends get. Suppose the first friend is x1, his second friend is x₂, and so on, all the way to the fifth friend. 

Then, x₁+x₂+x₃+x₄+x₅=20

Problem A) We can now use stars and bars, the number of non-negative solutions to this equation is (20+5-1) choose (5-1) which is 24 choose 4 which is 10626

Problem B) We can use stars and bars again, the number of positive integer solutions to the equation, x₁+x₂+x₃+x₄+x₅=20 is (20-1) choose (5-1) which is 3876 :)

If you want to learn more about stars and boxes, google is a good place.

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

\(x^2 - 4x = 4x - 20\)

\(x^2 - 8x + 20 = 0\)

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {8 \pm \sqrt{-16} \over 2}\)

\(x = {8 \pm 4i \over 2}\)

\(\color{brown}\boxed{x = 4 \pm 2i}\)

This problem is pretty easy if you know about discriminants.

Our equation is:

\(2x^2+5x+c=x^2-4x\) 

 First, move everything to the left side

\(x^2+9x+c=0\)

The discrimnant is the \(b^2-4ac\) (when the equation is \(ax^2+bx+c=0\)) part of the quadratic formula, basically

If the discriminant is greater than zero, there is 2 real solutions

If the discriminant is zero, there is one solution, a double root!

If the discriminant is less then zero, there is no real solutions

Using this we get: \(9^2-4c\) as the discrimant, which simplifies to \(81 - 4c\)

We want to get c as big as possible without this expression being negative, hence we want this to equal zero.

Solving \(81-4c=0\) 

 \(4c=81\)

\(c=81/4\)

I wrote thought but meant though, I'm too lazy to actually edit it though, so I'll just write this...

You should actually try to do these problems, they would help you a lot in the long run, because quadratics help in many math competitions...

But anyways, I'll still help:

We have:

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

When subtracting the x^2 from both sides we get:

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

Add 4x to both sides:

\(x^2+9x + c = 0\)

To have at least one real solution, and the largest... (Hmm, I actually don't know how to do this, but I have intuition!)

I just want you to think about this example for a moment (Solving by intuition again!)

93 * 95 = 8835

94 * 94 = 8836

Do you see something intriguing here? The closer the numbers are together, the greater their product is! (Try it!)

So, we would need two numbers that are close together, possibly equal, and then we will get the greatest product...

We find that these two numbers would have to add to 9, and if we want them to be equal, we use 4.5!

So 4.5 * 4.5 = 20.25, so our constant "c" would be 20.25!

We have to write this as a fraction thought, so:

20 1/4 = 81/4

(Hope my explanation was good, and if there was anything wrong please notify me!)

Honestly, I do think this is right, so could you try again a different way, or maybe they are asking for something more specific... 

@ImCool I'm trying to find your mistake gimme a sec... 

YESSS I FOUND IT (I'm so pleased with my self)

Vieta states that with roots r and s in a simple quadratic:

r+s = negative b

I'm pretty sure this is right based on my knowledge, so that's where you went wrong I think...

We have:

x/x-5 = 4/x-4

Multiply by (x-5) and by (x-4) to both sides

x(x-4) = 4(x-5)

x^2 - 4x = 4x - 20

Put the left side on the right...

x^2 - 8x + 20 = 0

We find factors of 20 that multiply to 20, but add to -8 (we realize factors must be negative...)

Instead, we can just put this into our quadratic formula:

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

So we have:

8 +/- root(64 - 80) }/2

8 +/- root(-16) }/2

8 +/- 4i

4 +/- 2i

So the answer should be 4 +/- 2i, note that the 4 is positive...

(If this answer is incorrect please notify me as I would be happy to help) (But you did say this is incorrect, so I don't know what's going wrong here...)