Answer:
• let f(x) be m:
• make x the subject of the function:
![{ \rm{8m = {x}^{3} + 128}} \\ \\ { \rm{ {x}^{3} = 8m - 128 }} \\ \\ { \rm{ {x}^{3} = 8(m - 16) }} \\ \\ { \rm{x = \sqrt[3]{8} \times \sqrt[3]{(m - 16)} }} \\ \\ { \rm{x = 2 \sqrt[3]{(m - 16)} }}](https://tex.z-dn.net/?f=%7B%20%5Crm%7B8m%20%3D%20%20%7Bx%7D%5E%7B3%7D%20%20%2B%20128%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7B%20%7Bx%7D%5E%7B3%7D%20%3D%208m%20-%20128%20%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7B%20%7Bx%7D%5E%7B3%7D%20%3D%208%28m%20-%2016%29%20%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7Bx%20%3D%20%20%5Csqrt%5B3%5D%7B8%7D%20%20%5Ctimes%20%20%5Csqrt%5B3%5D%7B%28m%20-%2016%29%7D%20%7D%7D%20%5C%5C%20%20%5C%5C%20%7B%20%5Crm%7Bx%20%3D%202%20%5Csqrt%5B3%5D%7B%28m%20-%2016%29%7D%20%7D%7D)
• therefore:
Answer:
4.5 % alloy
Step-by-step explanation:
Let x = amt of 30% alloy
The resulting total is to be 25 oz, therefore:
(25-x) = amt of 5% alloy:
A typical mixture equation:
.30x + .05(25-x) = .20(25)
.30x + 1.25 - .05x = 5 .30x - .05x = 5 - 1.25
.25x = 3.75
x = 3.75%2F.25
x = 15 oz of 30% alloy required
then
25-15 = 10 oz of 5% alloy:
Check solution
.30(15) + .05(10) = .20(25)
4.5 + .5 = 5
Answer:
32
Step-by-step explanation:
P=2(l+w)
so 2(11+5)=P
2(11+5)=32
Answer:66
Step-by-step explanation:
an=21+5n
ninth term is a9
a9=21+5x9
a9=21+45
a9=66
Answer: the answer would be c
Step-by-step explanation:
