Answer:
4 cookies
Step-by-step explanation:
Let the number of cookies he sold be c and that of brownies be b.
Assuming that brownies and cookies are the only type of baked goods sold,
b +c= 10 -----(1)
Amount of money received= $20
b(cost of brownie) +c(cost of cookie)= $20
2.50b +1.25c= 20 -----(2)
From (1): b= 10 -c -----(3)
Substitute (3) into (2):
2.50(10 -c) +1.25c= 20
Expand:
2.50(10) +2.50(-c) +1.25c= 20
25 -2.50c +1.25c= 20
-1.25c +25= 20
Being constants to 1 side:
-1.25c= 20 -25
-1.25c= -5
c= -5 ÷(-1.25)
c= 4
Thus, Joe sold 4 cookies.
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
The prime factorization of 51 is the product of prime numbers which results in 51. Thus, the prime factorization of 51 is 51 = 3 × 17.
Every possible combination of the letters SURE are going to be listed in alphabetical order. The permutation we want is RUSE which begins with the letter R and will come after every permutation that begins with E since it is the next alphabetically. We can first determine how many permutations begin with E.
Since we start with E, there are only three letters left to form the rest of the permutation. So 3! = 3*2*1 = 6 states that there are 6 permutations that can be made from the remaining three letters. So there will be 6 permutations that begin with E.
Using this same logic, we now know that there are 6 permutations that begin with the letter R. The letters USE are in reverse alphabetical order, which means that the word RUSE will appear as the last permutation that begins with R.
We know there are 6 permutations that begin with E, followed by 6 permutations that begin with R, making 12 total at this point. And since RUSE appears as the last permutation beginning we R, we know that RUSE shows up 12th.
