Set up the porportion: x/120= 40/100
Cross multiply: 100x=120 times 40
Hello,
Let's a=279-i with i=0 to 16
Let's b=312-i with i=0 to 16
Let's c=697-i with i=0 to 16
Let's d=708-i with i=0 to 16
I don't know how to solve this but
If n is the commun divider, it must be <√279=16.70...
Remainder must also be < n
We have to find the maximun of (a-i)∧(b-i)∧(c-i)∧(d-i) =11 (if i=4 or i=15 )
279=11*25+4
312=11*28+4
697=11*63+4
708=11*64+4
When i=15 ,n=11 but i must be<11
so only one solution n=11 and remainder=4
Answer:
(f+g)(3) = 347
Step-by-step explanation:
(f+g)(3) = f(3) +g(3) = (2·3 +210) +(2·3 +125) = 216 +131
(f+g)(3) = 347
____
The sum of two functions is the sum of their individual values.
Answer:
6 onuces per serving
Step-by-step explanation: