A. you would put in 7 for X and 3 for Y.
3 + 7 = 10
2(10) = 20
GP = 1/2 , 1/4 , 1/8 , 1/16
1st) the common ratio is (1/4 : 1/2 ) = 1/2, so r =1/2
2nd) the sum of a GP is:
S= a(1-rⁿ)/(1-r)
S=(1/2).[1-(1/2)⁴]/(1-1/2) = 15/16
Answer:d
Step-by-step explanation:
16-4=12 and 16+4=20
Answer:
-82/3
Step-by-step explanation:
first divide -8/9 by -2/3
-8/9 ➗ -2/3 =-8/9*3/-2 =4/3
multiply 4/3 by -41/2
4/3*-41/2 = -82/3
Let
x-------> the number of dinner
y-------> the number of lunch
we know that
-------> equation A
------> equation B
Substitute equation B in equation A
![8[y]+5y \leq 42](https://tex.z-dn.net/?f=8%5By%5D%2B5y%20%5Cleq%2042)



so
the greatest number of lunch is 

Hence
the greatest number of dinner is 
therefore
the greatest number of meals is

<u>the answer is</u>
