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Alright, let's see what we can do here...

\(x^2 - 25x + 80 = x^2 - 16x + 44 \)

Subtract x^2 from both sides

\(36 = 9x  \)

\(x = 4 \)

So I got x = 4

For the other part of your question, I don't really understand what it is asking, so you can figure that out...

(Please tell me if what I did so far was incorrect, and if you could clarify the other part then I'd be happy to help)

The solutions are actually 3 \pm 2i.

Sorry, i tried typing in -4\pm2i and it said it was incorrect. Thank you tho for your effort in helping me!

Ok lets see...

alright lets make these look better.

\(\frac{x(x-4)}{(x-5)(x-4)}=\frac{4(x-5)}{(x-4)(x-5)}\)

alright more simplifying:

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

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

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

Oh yay! A quadratic equation! Now time to solve!

Ok so:

\(x+y\:=\:-8,\:xy=20\)

\(x=-8-y\)

Substitute this into xy = 20 and you will get

\(\begin{pmatrix}x=-4+2i,\:&y=-4-2i\\ x=-4-2i,\:&y=-4+2i\end{pmatrix}\)

so x is NOT 4±2i but it is -4\pm2i so yeah!

If anyone can check my work that would be great

Next time the image would be greatly appreciated, lets see what i can do..

What do you mean by finding the equation of the medians? I think it is how to find the medians

ok so find the midpoint of the line segments with this:

\((\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\)

for part b thou, they intersect at \(\left(2.66666666666,2.3333333333\right)\)

Its correct, thank you so much!

For it to be the square of a binomial, it has to be in the form:

\((r + s)^2 = 4x^2 + 14x + c\)

We find that the first term squared = 4x^2

\(r^2 = 4x^2\)

\(r = 2x\)

\((2x + s)^2 = 4x^2 + 14x + c\)

We know the second term is in the form:

\(4(x)(r) = 14x\)

\(4r = 14\)

\(r = 3.5\)

So our binomial is in the form:

\((2x+3.5)(2x+3.5) \)

So our constant "c" would be:

\(c = 3.5^2 \)

\(c = 12.25\)

(If this answer is wrong please tell me, as I would be more than happy to fix it)

I solved it a different way and got a different answer:

So we have the 4th square root of 7 over the 3rd square root of 7 is equal to 7 raised to what power...

Let's write this as an expression, with x being the exponent of 7:

\((\sqrt[4]{7})/(\sqrt[3]{7}) = 7^x\)

We simplify:
\((7^{1/4})*(7^{-1/3}) = 7^x\)

We know when multiplying numbers with same bases, we add exponents so:

\(7^{1/4-1/3}=7^x\)

\(7^{3/12-4/12}=7^x\)

\(7^{-1/12}=7^x\)

This simplifies to:

\(x = -1/12\)

So the exponent is:

-1/12

(I advise you to read my work, because I could be wrong)

When looking at the lengths of the sides of the triangle, you realize you can use a certain formula and ignore all the random lines inside the triangle. Heron's formula tells you the area of a triangle given its 3 side lengths. 

By heron's formula (which I will not prove and you can look it up on your own), the area of the triangle is \(\sqrt{42(42-36)(42-22)(42-26)} = \sqrt{42*8*20*16} = 32\sqrt{105}\)

I'm thinking to solve it a different way...

We have:

\(81^n=9^{27} \cdot 27^9\)

We know 81 is 3^4

\((3^4)^n = 9^{27} \cdot 27^9\)

So the left side simplifies to 

\(3^{4n} = 9^{27} \cdot 27^9\)

We know 9 is 3^2 and 27 is 3^3 so we use that information into our equation:

\(3^{4n} = (3^2)^{27} \cdot (3^3)^9\)

So the right side simplifies to:

\(3^{4n} = (3^{54}) \cdot (3^{27})\)

Then, we know that if we have the same base when multiplying, we add the exponents so:

\(3^{4n} = (3^{81})\)

So:

\(4n = 81 \)

Divide by 4 to both sides:

\(n = 81/4 = 20.25\)

This is (I believe) the standard and faster way to solve it in a real competition. 

Why are you having problems with this simple question? If your calculator display isn't large enough, then use the calculator on this site!

81^n=9^27 * 27^9?

1 - With a calculator, calculate the right-hand side first: 9^27 * 27^9 =443426488 2430377699 4824963061 9149892803

2 - Take the log (base 10) of the above number: log(443426488 2430377699 4824963061 9149892803)

3 - Log = 38.6468216322 (you only need the log to 10 decimal places).

4 - Take the log of 81 to 10 decimal places: log(81) = 1.9084850189

5 - Divide the log in (3) above by the log in (4) above: 38.6468216322 /  1.9084850189=20.25

6 - Therefore: n = 20.25.  And that is it!

There are 7,244 ways.

thx but i need solution is okay if you provide me with a solution?

(a) There are 20 ways.

(b) There are 32 ways.

There are 480 ways.

The correct value of n is 16/3.

wbout the big point theorem :( make the points really big, and they are on the same line. :) especially when  you draw a diagram

The center of the circle  (x + 4)² + (y - 3)²  =  9  is the point  (-4, 3).

Moving the point  (-4, 3)  to the point  (6, 3)  takes 10  units.

The vertices of a triangle are, by definition, noncollinear; therefore, you cannot prove that they are collinear.

Counting the ways:

1) You can have 10 zeros ==  1 way

2) You can have 1 one and 9 zeros == 10 ways

3) You can have 1 two and 9 zeros == 10 ways

4) You can have 2 ones and 8 zeros:

     If you start with a 1: 9 ways to place the other one.

     If you start with 0 1: 8 ways to place the other one.

     If you start with 0 0 1: 7 ways to place the other one.

     If you start with 0 0 0 1: 6 ways to place the other one.

     ... continue with this and add all the ways.

Instead of posting it on here right away, try checking previous ones, if you scroll back a few days, there was someone else who asked the same question, so go look there.

I'm so disappointed. You didn't show us the "grid below". There's nothing below. Please reread your question before posting it on here, so you can check for mistakes.

Note that all 6 digit numbers work, so there are \(6 \times 6 \times 6 \times 6 \times 6 \times 6 = 46,656 \)numbers

Now, for the 5 digit numbers, there are \(6^5 = 7776\) numbers, but we need to count for the numbers that are less than 23,000.

If the first digit is a 1, there are \(6^4 = 1296\) numbers that don't work

If the first two digits are 21, there are \(6^3 = 216\) numbers that don't work

LIkewise, if the first two digits are 22, there are 216 numbers that don't work

So, there are \(46,656 + 7776 - 1296 - 216 - 216 = \color{brown}\boxed{52704}\)