Given that the alphabet soup contains the vowels (A, E, I, O, U) with each of the vowels appearing five times.
This means that the total number of letters in the bowel soup is 25 letters.
The number of five-letter words that can be formed from the bowel of 25 letters assuming there are no restrictions is given by
Answer:
-1.5x + 21.5 (or (43-3x)/2)
Step-by-step explanation:
(-5/6x + 51/2) - (2/3x + 4)
<em>First, find the opposite of 2/3x + 4 because of the negative sign.</em>
-5/6x + 51/2 - 4/6x - 4
<em>Now combine -5/6x and -4/6x to get -9/6x or -3/2x</em>
-3/2x + 51/2 - 4
<em>Convert 4 into 8/2 since 51/2 is a fraction.</em>
-3/2x + 51/2 - 8/2
<em>Since the denominators of 51/2 and -8/2 are the same, just subtract 51 and 8 to get 43.</em>
-3/2x + 43/2
<em>-3/2x and 43/2 have the same denominator (2), so you can just put it like this (It's up to you!):</em>
(43-3x)/2
The answer to (-5/6 + 51/2) - (2/3x + 4) is -3/2x + 43/2 (or (43-3x)/2)
<em>Or you could put it like this (I converted both -3/2 and 43/2 to a decimal.)</em>
-1.5x + 21.5
8 x 6 = # of slices
If this is not what you wanted, let me know.
24/200×100/100 = 12/100 = 12%
