Sub (x+1) for x
f(x+1)=2(x+1)^2+3
f(x+1)=2(x^2+2x+1)+3
f(x+1)=2x^2+4x+2+3
f(x+1)=2x^2+4x+5
D is answer
Co linear points have same slope
<span>Points (1,3), (-4,7), and (-29,K)
m1,2=7-3/-4-1=-4/5
m2,3=K-7/-29+4
m2,3=K-7/-25
-4/5=K-7/-25
4/5=K-7/25
4=K-7/5
20=K-7
K=27
</span>
Ohhhh nasty ! What a delightful little problem !
The first card can be any one of the 52 in the deck. For each one ...
The second card can be any one of the 39 in the other 3 suits. For each one ...
The third card can be any one of the 26 in the other 2 suits. For each one ...
The fourth card can be any one of the 13 in the last suit.
Total possible ways to draw them = (52 x 39 x 26 x 13) = 685,464 ways.
But wait ! That's not the answer yet.
Once you have the 4 cards in your hand, you can arrange them
in (4 x 3 x 2 x 1) = 24 different arrangements. That tells you that
the same hand could have been drawn in 24 different ways. So
the number of different 4-card hands is only ...
(685,464) / (24) = <em>28,561 hands</em>.
I love it !
