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Rocco is a member of the Aviation Club, which has 12 members.

The members need to choose a 3-person Helicopter Committee and a 4-person Glider Committee.

A member can serve on one of the committees or both. Rocco refuses to serve on both (but he can serve on one of the committees).

In how many ways can they form the committees?

If rocco is on the helicopter committee then

there are   11C2*11C4 = 55*330 = 18150 ways

If rocco is on the Glider committee committee then

there are   11C3*11C3 = 165*165 = 27225 ways

If rocco is not on either committee then

there are   11C3*11C3 = 165*330 = 54450 ways

18150+27225+54450 = 99825 ways

Abraham lincon was our 16th president

our???

He was never president of my country.

Since you have copied this from another site you need to acknowledge where it came from.

Hello Guest!

\(\angle ACD=\angle ABC =x\\ \angle BDC=\angle BCD=y \)

so:

2y+x=180°

x=180°-2y

x=180°-2y

48°+(180°-y)+x=180°

48°+(180°-y)+(180°-2y)=180°

-3y=(180-180-180-48)°

-3y=-228°

y=76°

x=180°-2*76°

x=28°

\(\angle ABC=28°\)

  !

Firstly neither are written properly Person A gets 3% per annum (yearly)  and Person B gets 4% per annum (yearly) 

BOTH STATED interest rates are per annum.

A will have        2700*(1.03)^n     where n is in years

B will have       2300*(1+0.04/12)^m     where m is in months

Now you can work them out

The angles in each of those 3 similar tirangles are   66,60 and 54  

The 3 triangles are similar so their angles are 66,60 and 54

angle MNO must be 54 so that MNO can be 60

which means that angle NRO must be 60 degrees

It would be better to do  formal solution though.

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NOTE: I am editing because it mostly disappeared when I published. How annoying.

I have not seen that compilation error for a long time.

\(\frac{n}{2}((2(\frac{3}{4})+(n-1)(\frac{3}{8}))-\frac{\frac{3}{4}((\frac{1}{2})^n-1)}{\frac{1}{2}-1}\ge200\\ \frac{n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))-\frac{\frac{3}{4}((\frac{1}{2})^n-1)}{-\frac{1}{2}}\ge200\\ \frac{n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))+2*\frac{3}{4}((\frac{1}{2})^n-1)\ge200\\ \frac{n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))+\frac{3}{2}((\frac{1}{2})^n-1)\ge200\\ \frac{16n}{2}(\frac{3}{2}+(n-1)(\frac{3}{8}))+\frac{16*3}{2}((\frac{1}{2})^n-1)\ge3200\\ \frac{8n}{1}(\frac{3}{2}+(n-1)(\frac{3}{8}))+\frac{8*3}{1}((\frac{1}{2})^n-1)\ge3200\\ 12n+3n(n-1)+24*(\frac{1}{2})^n-24\ge3200\\ 12n+3n^2-3n+24*(\frac{1}{2})^n\ge3224\\ 9n+3n^2+24*(\frac{1}{2})^n\ge3224\\ 24*(\frac{1}{2})^n\ge -3n^2-9n+3224\\ \frac{1}{2^n}\ge \frac{ -3n^2-9n+3224}{24}\\ \)

Thanks 

Yes thanks guys and girls 

....

f(x)=|x|

g(x)= -f(x)          or      g(x)= -|x|

They are probably kids in the same class. 

It would be good if  we could do something like that but this site is not organised enough to manage it.

The kids would not look anyway. 

....

They are different graphs 

Thank you, SVS!

You remove one from it.

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1.Do the prime factorization of the number.Ex:a^p b^q c^r
2.Total no.of factors=(p+1)*(q+1)*(r+1)

Similarly here we have to do the prime factorization of the number 110n^3

prime factorization=110n^3=2*5*11*n^3
no.of factors=110 (already given)

Now we have to create a analogy in the system

no .of factors = 110=2*5*11

this means
1>respective powers of the prime factors will be one less than the above so they will be (1,4,10)
2>there will be total 3 factors ,so n much contain prime factors of one of the (2,5,11)

Now we have the full liberty to assign any power among above powers to any prime factors , but looking at the options below, we should not raise 5and 11 to the higher powers.

So considering 2^10 5^4 11^1 as one such combination because it will yield a number smaller in value and in horizon near to the answer choices.

So this means that 110n^3=2^10*5^4*11^1==> n=2^3*5

So 81*n^4=(3*n)^4==>2^12*5^4*3^4 (prime factorization)

Sum of factors ==>(12+1)*(4+1)*(4+1)==>325

Good Iowa is freeeeezing

You Got it Lol

....

This question was just answered in the previous forum 

G(x) = |x|-2 in the frst one
g(x) = -|x| in the second one

cool,did you create this website?

2x + 3 = x + x + 3

1 / 1005
1 / 1003
1 / 502
2 / 503
1 / 168
2 / 169
3 / 170
This continues until you get: 166 / 333 + 1
165 / 332
166 / 333
1 / 2
2 / 3
3 / 4 - This continues until you get: 2004 / 2005
2001 / 2002
2002 / 2003
2003 / 2004
2004 / 2005
TOTAL TERMS = 2,174