Answer:
there will be 19 $100 bills
and 11 $10 bills
Step-by-step explanation:
Answer:
Step-by-step explanation:
did u find the answer
The only expressions that are correctly factored are;
A) 16a⁵ - 20a³ = 4a³(4a² - 5)
B) 24a⁴ + 18 = 6(4a⁴ + 3)
C) 12a³ + 8a = 4a(3a² + 2)
D) 30a⁶ - 24a² = 3a²(10a⁴ - 8)
<h3>How to factorize equations?</h3>
1) 16a⁵ - 20a³
To factorize this, we will have to get out the common factor first. The common factor is 4a³. Thus, we now have;
4a³(4a² - 5)
2) 24a⁴ + 18
To factorize this, we will have to get out the common factor first. The common factor is 6. Thus, we now have;
6(4a⁴ + 3)
3) 12a³ + 8a
To factorize this, we will have to get out the common factor first. The common factor is 4a. Thus, we now have;
4a(3a² + 2)
4) 30a⁶ - 24a²
To factorize this, we will have to get out the common factor first. The common factor is 3a². Thus, we now have;
3a²(10a⁴ - 8)
Read more about factorization of equations at; brainly.com/question/723406
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The volume of a pentagonal prism that has a base area of 5.16 cm^2 and a height of 9cm is 46.44cm^3
Have fun!
9514 1404 393
Answer:
- 85 ones
- 70 fives
- 15 tens
- 2 fifties
Step-by-step explanation:
Let a, b, c, d represent the numbers of $1, $5, $10, and $50 bills, respectively. The problem statement tells us ...
a +5b +10c +50d = 685 . . . . . total amount of cash
b = 10 +4c . . . . . . . . . . . . the number of fives is 10 more than 4 times tens
a = b+c . . . . . . . . . . . as many ones as fives and tens combined
d = 2 . . . . . . . . . . the register contained 2 fifty-dollar bills
__
Substituting for 'a', then for 'b', we have ...
(b+c) +5b +10c +50d = 685
6(10 +4c) +11c +50d = 685
60 +35c +50d = 685
Substituting d=2 and subtracting 160 gives ...
35c = 525
c = 15
b = 10 +4c = 10 +4(15) = 70
a = b+c = 70 +15 = 85
The register contained ...
- 85 ones
- 70 fives
- 15 tens
- 2 fifties
