40$. 7hrs
X$. 1hr
X=40•1/7
X~5.71 $ per hour
did i hear help i will help you soooooooo oooooooooo
I believe that it would be a-21/49
<span>let
x = acres of when
y = acres of rye
Maximize
z = 500x + 300y
subject to
x >= 3
x <= 15
y >= 1
y <= 7
x + y <= 20
bounded
both
(3, 1)
(15, 1)
(15, 5)
(13, 7)
(7, 3)
sub points into your max equation
(3, 1) = 1500 + 300 = 1800
(15, 1) = 7500 + 300 = 7800
(15, 5) = 7500 + 1500 = 9000
(13, 7) = 6500 + 2100 = 8600
(7, 3) = 3500 + 900 = 4400
max profit of $9,000 is achieved when 15 acres of wheat and 5 acres of rye are planted and sold</span>
Answer:
3.
Step-by-step explanation:
This is a geometric series so the sum is:
a1 * r^n - 1 / (r - 1)
= 1 * (2^101 -1) / (2-1)
= 2^101 - 1.
Find the remainder when 2^101 is divided by 7:
Note that 101 = 14*7 + 3 so
2^101 = 2^(7*14 + 3) = 2^3 * (2^14)^7 = 8 * (2^14)^7.
By Fermat's Little Theorem (2^14) ^ 7 = 2^14 mod 7 = 4^7 mod 7.
So 2^101 mod 7 = (8 * 4^7) mod 7
= (8 * 4) mod 7
= 32 mod 7
= 4 = the remainder when 2^101 is divided by 7.
So the remainder when 2^101- 1 is divided by 7 is 4 - 1 = 3..
