We have been given in a cohort of 35 graduating students, there are three different prizes to be awarded. We are asked that in how many different ways could the prizes be awarded, if no student can receive more than one prize.
To solve this problem we will use permutations.
We know that formula for permutations is given as
On substituting the given values in the formula we get,
Therefore, there are 39270 ways in which prizes can be awarded.
So for this, you will be doing two different multiplications: 3 x 4 and √8 x √3.
3 x 4 = 12
√8 x √3 = √24
Now our result is 12√24, however, this can be simplified. Using the product rule of radicals (√ab = √a x √b), our simplification is such:
12√24 = 12√(8 x 3) = 12√(4 x 2 x 3) = 2 x 12√(2 x 3) = 24√6
In short, the answer is 24√6, or the first option.
Answer:
8w² < 4w(150-w)
Step-by-step explanation:
Square area of living : w · w = w²
Money spent : 8 · w²
Square area of artichokes : (150 - w) · w
Money earned : 4 · w · (150 - w)
Julia manages to save some money every week. That means that the money earned is bigger than the money spent ( the money spent is less than the money earned)
8 · w² < 4 · w · (150 - w)
Answers:Part A: The value of x is 0.Part B: X can be any real number.
In Part A, you have to first evaluate 7^2. This is 49. Now, write the equation 49^x = 1. We know that if you raise any number to 0, then the answer is 1.
In Part B, you have to first evaluate 7^0, that is 1. Now, we have the equation 1^x = 1. In this case, 1 raised to any exponent is still only 1. Imagine 1^17, this would be 1 times itself 17 times or just 1.
Therefore any number will work in Part B.
Answer:
is the number 4
so 13/16 3/16
Step-by-step explanation:
