Case a: A student can receive any number of awards
Let's count our choices: the first award can go to any of the 20 students. So we have 20 choices. The second awards can also go to any of the 20 students. So we have 20*20 choices for the first two awards. Similarly, we have 20*20*20 choices for the first three awards, and so on.
So, there are possible ways to give the awards, if a student can receive as many awards as possible.
Case b: A student can receive only one awards
This will be very similar to the previous case, but with a minor restriction: as before, we have 20 choices for the first award, because it can go to any of the 20 students.
But when it comes to the second award, we only have 19 choices, because we can't give it to the student who already won the first award.
Similarly, we can give the third award to one of the 18 remaining students, because we can't give it to the students who already won the first or second award.
So, in the end, we have
ways of awarding the students, if a student can win only one award.
Answer:
D is correct
Step-by-step explanation:
because 7×2=14 and 7 times as 2 is 14
please mark me as brainliest answer
Answer:
16
Step-by-step explanation: parenthesis first (2+2)=4
then division 8/2=4
then multiply 4x4=16
What’s the question??? And the answer choices ??
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.