a^4+2a+8 because if you combine like terms that is what you get
Since sin(2x)=2sinxcosx, we can plug that in to get sin(4x)=2sin(2x)cos(2x)=2*2sinxcosxcos(2x)=4sinxcosxcos(2x). Since cos(2x) = cos^2x-sin^2x, we plug that in. In addition, cos4x=cos^2(2x)-sin^2(2x). Next, since cos^2x=(1+cos(2x))/2 and sin^2x= (1-cos(2x))/2, we plug those in to end up with 4sinxcosxcos(2x)-((1+cos(2x))/2-(1-cos(2x))/2)
=4sinxcosxcos(2x)-(2cos(2x)/2)=4sinxcosxcos(2x)-cos(2x)
=cos(2x)*(4sinxcosx-1). Since sinxcosx=sin(2x), we plug that back in to end up with cos(2x)*(4sin(2x)-1)
You have a 1 in 50 chance (ie A)
Explanation:
Let's start with the first bag. In it sits 5 numbers. So, you have a 1 in 5 chance that the number picked will be yours.
Now, let's consider the second bag. You have 10 different numbers and only one of yours may be picked. Hence, you have a 1 in 10 chance that yours will get picked.
Since we want them simultaneously, we will need to multiply the two probabilities together, yielding a 1 in 50 chance.
Y=2x+10
That is the eqasion, 2(it is the cost per card) and multpy be how many cards you want, then add 10 as base fee
The answer to the question is C)$33 because you do 6•2=12 and 7•3=21 and you add them which is $33 but you don’t add $10 and $13
