The valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b
<h3>How to determine the valid exclusion?</h3>
The expression is given as:
8ab^2x/4a^2b - 8ab^2
Set the denominator to 0
4a^2b - 8ab^2 = 0
Divide through by 4ab
a - 2b = 0
Add 2b to both sides
a = 2b
Hence, the valid exclusion of the algebraic fraction is (c) a =0, b =0, a =2b
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1006.103. It is just breaking apart the final answer into smaller numbers
The answer is C . you have to solve it like that
Printed pages in 11 minutes = 33 pages
Printed pages in a minute = 33/11
= 3 pages/minute
Answer:
the three consecutive odd integers are 9, 11, and 13.
Step-by-step explanation:
Let's say that the first odd integer is x. The second consecutive odd integer would have to be x+2. (It would not be x+1 because that would result in an even integer. The sum of any two odd numbers is even.) The third consecutive odd integer would be (x+2) +2 or x+4.
The sum of the first, twice the second, and three times the third can be written as:
x+2(x+2)+3(x+4)
This equals 70. We can now distribute and solve for x:
x+2(x+2)+3(x+4)=70
x+2x+4+3x+12=70(distribute)
6x+16=70(combine like terms)
6x=54(subtract 16 from both sides)
x=9(divide by 9)
Thus, the three consecutive odd integers are 9, 11, and 13
hope this helps :)
