Please explain how to solve this problem A nut store normally sells cashews for​ $4.00 per pound and peanuts for​ $1.50 per pound. But at the end of the month the peanuts had not sold​ well, so, in order to sell 80 pounds of​ peanuts, the manager decided to mix the 80 pounds of peanuts with some cashews and sell the mixture for $2.00 per pound. How many pounds of cashews should be mixed with the peanuts to ensure no change in the​ revenue?

The manager should mix _ pounds of cashews with the peanuts.

9514 1404 393

  20 pounds

Step-by-step explanation:

Often, mixture problems are worked by writing two equations: one for the total quantity of constituents in the mix, and one for the total price of the mix. When these equations are solved by substitution, the first step is to write one quantity in terms of the other.

You can save an equation and a step by using one variable for quantity. It usually works best to let the variable represent the highest-cost (or most concentrated) contributor. Here, that is the quantity the problem asks for, so ...

Let c represent the number of pounds of cashews needed to be mixed with the peanuts. Then the revenue from selling the mix is ...

  80($1.50) +c($4.00) = (80+c)($2.00)

  120 +4c = 160 +2c . . . . . divide by $, simplify

  2c = 40 . . . . . . . . . . . . . subtract 120+2c from both sides

  c = 20 . . . . . . . . . . . . . divide by 2

The manager should mix 20 pounds of cashews with the 80 pounds of peanuts.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

$9.41

step-by-step explanation: