Answer:
Use 49 ounces of the 14% allow and 41 ounces of the 23% alloy.
Step-by-step explanation:
Each ounce of the 14% copper contains 0.14 ounce of pure copper.
Each ounce of the 23% copper contains 0.23 ounce of pure copper.
Each ounce of the 18.1% copper contains 0.181 ounce of pure copper.
Use x ounces of the 14% and y ounces of the 23% to make 90 ounces of 18.1% alloy.
x+y = 90
y = 90-x
0.14x + 0.23y = 0.181·90
0.14x + 0.23(90-x) = 16.29
0.14x + 20.7 - 0.23x = 16.29
-0.09x + 20.7 = 16.29
4.41 = 0.09x
x = 49
y = 90-x = 41
Use 49 ounces of the 14% allow and 41 ounces of the 23% alloy.
Neither of them are correct. When you plug in both of the fractions into the equation, you get answers like 2/9 (from plugging in 1/3) and 1/9 (from plugging in 1/6). None of them are equal to 9.
Answer:
A
Step-by-step explanation:
i think thats the answer
Answer:
A) JK and AB =
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So, =》 
=》 
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B) AABC
![[tex] \boxed{( \theta) = {1}{( \theta)} }](https://tex.z-dn.net/?f=%5Btex%5D%20%5Cboxed%7B%28%20%5Ctheta%29%20%3D%20%7B1%7D%7B%28%20%5Ctheta%29%7D%20%7D)
So, 
C) AABM:
![[tex] e \fbox{: }](https://tex.z-dn.net/?f=%5Btex%5D%20e%20%5Cfbox%7B%3A%20%7D)
The value of k for which 
is 
ans[/tex]
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