Blaine is selling tickets to a rock concert. tickets in the front section cost $60 each. tickets in the back section cost $40 each. blaine has 55 tickets left to sell, and to meet his sales quota he needs to sell more than $600 in tickets from the remaining tickets.
blaine makes a profit of $40 on every ticket for the front section. he makes a profit of $25 on every ticket for the back section. what’s the maximum profit blaine can make from selling his remaining tickets?
it has to be two system of inequalities.

hello from mrbilldoesmath!

answer:

a(n) - a(n-1) -3

discussion:

this ones a bit interesting. given a(n) = 5 - 3n consider this:

a(n-1) = 5 - 3(n-1)

          = (5 - 3n) + 3             => but a(n) = 5 - 3n, the first term

          = a(n)   + 3                 => subtract 3 from both sides

a(n-1) -3 = a(n)  

or

a(n) - a(n-1) -3

you,

mrb

for any of you who need the translation here it is

step-by-step explanation:

find the angular size of the equatorial radius of the earth, if the horizontal parallax of the sun

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743