Imcool

I guess Heron's formula could work.... oh boy. Gimme a sec. 

MENTION: I set AE as 2x so its easier.

\(\sqrt{(4+x)(4-x)(1+x)(x-1)}+\sqrt{(3.5+x)(1.5+x)(x-1.5)(3.5-x)} = \sqrt{18.75}\)

\(x=\sqrt{\frac{2.25322E15}{1.0E15}},x=\sqrt{\frac{-2.25322E15}{1.0E15}}\)

\(ans=\frac{\sqrt{\frac{2.25322E15}{1.0E15}}}{2}\)

ahem....

this is 100% wrong..

approx. \(\frac{\sqrt{112661}}{100\sqrt{5}}\)

I'll try my best on this one, I'm not particularly good with this topic.

So the first Ace can appear anywhere, so 52 choices, times 4 since it could be any of the 4 aces.

The ace adjacent to it has 2 choices, left or right, and has 3 choices on which type of ace.

The probability of this happening is \(\frac{52\cdot4\cdot2\cdot3}{52\cdot52\cdot2\cdot51}\) which will end up being \(\frac{1}{221}\)

so the expected number of times is 0.

Ive googled this question. Electric Pavlov also answered this, and he got no answer as well. Check with your teacher, or the website, or whoever made this question.

you are x, your brother is y

\(x-5 = y+5\)

\(x+10 = 2(y-10)\)

now we solve

\(x+10 = 2y-20\)

\(x = 2y-30\)

\(x = y+10\)

\(2y-30 = y+10\)

\(y = 40\)

your brother has 40 dollars right now.

Alright, we set the largest of the 7 number to be x.

\(x-6+x-5+x-4+x-3+x-2+x-1+x = 214x\)

\(7x-21=214x\)

\(-21 = 207x\)

\(x = -21/207\)

that is not possible since they have to be integers.

hmmmm. Is this a trick question?

NOTE: notice that \(\frac{2x+3}{2x+3} = 1\)

a. since both are equal to one, any number we plug in will just be 1.

b. \(x=-\dfrac32\)for this, the denominator will equal zero when you plug it in.

c. No other real numbers can make the denominator 0, therefor all others work.

So.

Since it is a bisector, DC will be 10. Since Angle C is 15 degrees, we can write DE as

\(10\cdot tan(15^\circ)\)

That is equal to

\(10\cdot \frac{\sqrt{3}-1}{\sqrt{3}+1}\)

which is equal to 

\(20-10\sqrt{3}\)

Since

\(BE = 2\sqrt{200-5\sqrt{3}}\)

Im probably wrong, so anyone please correct me.