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I feel like step one is wrong.

The problem was, 

\(\frac{7}{9}-\frac{3}{8} \)

and step one multiplys them, which is incorrect.

The step reads the problem wrong.

Step 4 is correct however.

21/72 simplifys to 

7/24.

Your Welcome (: 

XD The Face! 

(: (: (: (: (: (: 

are yousure thats right or is 7 square root 2 

What are you expanding? Is it (x + 1/x)^6 ? I can't read your LaTex !.

oof ok.

ive got like five sisters and im the only boy so... figure

(5 +2) / y = y/2

7 / y = y / 2      cross myltiply

y^2 = 14

y = sqrt(14)

no i thought i did but then i got square root of 10 so no i don't get it still 

sorry can you explain it further plz 

How do you pay neg? Just curious.

Inverse of what?

Inverse in what situation?

Where are you typing this?

Please Clarify.

m∠ABD  =  m∠CBA  because they are the same angle

m∠BDA  =  m∠BAC  because they are both right angles.

So by the AA similarity theorem, triangle ABD is similar to triangle CBA

Do you see what to do next?

Please...

No comment.

And how do you know what the girls learn about anyways?

Really? Thank you guys!

m∠ABD + m∠CBD  =  90°

m∠ABD + m∠BAD  =  90°

So  m∠CBD =  m∠BAD

And  m∠BDA  =  m∠CDB  because they are both right angles.

So by the AA similarity theorem,  △ABD  ~ △BCD

BD / CD  =  AD / BD

x / 10  =  4 / x

x² / 10  =  4

x²  =  40

x  =  sqrt( 40 )

x  =  2 sqrt( 10 )

only if you post it

9^2010 mod 17 = 13

I belive its (C)

ok thank you 

\(N=12+15+25 = 52\\ P[70c]=\dfrac{12}{52}=\dfrac{3}{13}\\ P[40c]=\dfrac{15}{52}=\dfrac{5}{13}\\ P[30c]=\dfrac{25}{52}\\~\\ E=70 \dfrac{3}{13}+40 \dfrac{5}{13}+30\dfrac{25}{52}-50 = -\dfrac{205}{26}\approx -\$0.08\)

The answer is actually 34.

There are 35 different combinations of coins but two of them produce 40 cents.

I'd upload the Mathematica sheet but images seem to be broken atm.

here's a link to a pic of it

\(p_n = a\cdot n\\ \sum \limits_{n=1}^8p_n=432\\ a\sum \limits_{n=1}^8 n = 432\\ a \dfrac{8(8+1)}{2} = 432\\ 36a=432\\ a=12\\ 8\cdot 12 = 96~\text{pages}\)

Can you help me with another one like this one

\(\text{well we can get a few equations using the Pythagorean theorem}\\ \text{Let the left hand unlabeled leg be $a$ and the right hand one be $b$}\\ x^2+4^2=a^2\\ x^2+16=a^2\\~\\ a^2+b^2=(10+4)^2= 14^2=196\\~\\ x^2+10^2=b^2\\ x^2+100=b^2\)

\(\text{We can add the first and third equations}\\ x^2+16+x^2+100=a^2+b^2\\ 2x^2 + 116 = a^2+b^2\\~\\ \text{and note this equals the left hand side of the second equation}\\ 2x^2+116 = 196\\~\\ 2x^2=80\\ x^2=40\\ x=\sqrt{40}=2\sqrt{10}\\~\\ \text{choice B}\)

deleted.................  I am not sure about my answer