TaliaArticula

Hey there, Guest!

So, the equation is \(\left(a^b\right)^a-\left(b^a\right)^b\ \) and we'll substitute properly.

\(\left(2^3\right)^2-\left(3^2\right)^3\)

\(8^2−(3^2)^3\)

\(64−(3^2)^3\)

\(64−9^3\)

\(64−729\)

= −665

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

Yes, you're right, let me show you how:

The distance of the line x+y = k from the centre of the given circle is k/\(\sqrt{2}\)And the circle's radius is \(\sqrt{k}\). Therefore the tangency is (k/\(\sqrt{2}\)) of which equals \(\sqrt{k}\), which is equal to 2.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

I might've missed a few permutations that I'll find in the future, but for the moment, I'm pretty sure that the answer is 35.

Here's why:

4 0 0 0 x4 boxes

3 1 0 0 x12 permutations amongst the four boxes

2 2 0 0 x6

2 I 1 0 x 12

1 1 1 1

6 + 4 + 12 + 12 + 1 = 35.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

It appears that the question that you asked has already been answered here: https://web2.0calc.com/questions/more-hw-please-help.

In case if you're one of the people online of which don't like links, this is what it says:

In triangle PQR, PQ=13, QR=14, and PR=15. Let M be the midpoint of QR. Find PM.

Let angle PRQ be  >β<

1)   Let's find β  first:             cos(β) = (15² + 14² - 13²) / 2*15*14       β = 53.13°

2)   Now we can find  PM      (PM)² = 7² + 15² - 2*7*15 * cos(β)    

                                                 PM = 12.166     

Answerer: Dragan

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

It looks like CPhill already answered that question right here: https://web2.0calc.com/questions/mathematics_66.

Hope this helped!? XD

( ﾟдﾟ)つ Bye

Hey there, Guest!

It looks like CPhill already answered that question right here: https://web2.0calc.com/questions/mathematics_66

Hope this helped!? XD

( ﾟдﾟ)つ Bye

Hey there, Guest!

So, let's go through this problem step-by-step:

First, let's format it like this: \(\frac{x+1/x}{x-2/x}=\frac{5}{4}\)

\(\frac{x^2+1}{x^2-2}=\frac{5}{4}\)

Step 1: Multiply both sides by \(x^2-2\).

\(x^2+1=\frac{5}{4}x^2+\frac{-5}{2}\)

\(x^2+1-\frac{5}{4}x^2=\frac{5}{4}x^2+\frac{-5}{2}-\frac{5}{4}x^2\) (Subtract \(5/4x^2\) from both sides.)

\(-\frac{1}{4}x^2+1=\frac{-5}{2}\)

\(-\frac{1}{4}x^2+1-1=-\frac{5}{2}-1\) (Subtract from both sides.)

\(\frac{-\frac{1}{4}x^2}{-1/4}=\frac{\frac{-7}{2}}{-1/4}\)

\(x^2=14\)

Therefore \(x=\sqrt{14}\) and \(-\sqrt{14}\).

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Mnkmnk!

So I'm not quite sure if this can help, but here it is: https://web2.0calc.com/questions/math_83560.

And I also understand that I'm not @CPhill, so apologies about that.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Kirstenday51704!

So, let's do this:

3) (x+1)(x−6)

4) 2(x+2)(x−2)

5) (x+2)(x+10)

6) (3n−7)(n−9)

7) (x^2+4)(x+2)(x−2)

8) (2x+15)(x−1)

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

A pentagon has 5 sides.

Hope this helped! :)

( ﾟдﾟ)つ Bye