Solve this application problem using a system of equations: Dan and June mix two kinds of feed for pedigreed dogs. They wish to make 70 pounds of feed worth $0.30 per pound by mixing Feed A worth $0.26 per pound with Feed B worth $0.40 per pound. How many pounds of the cheaper kind should they use in the mix

50 pounds

Step-by-step explanation:

Dan and june mix two kind of feed for pedigreed dogs

Feed A worth is $0.26 per pound

Feed B worth is $0.40 per pound

Let x represent the cheaper amount of feed and y the costlier type of feed

x+y= 70equation 1

0.26x + 0.40y= 0.30×70

0.26x + 0.40y= 21equation 2

From equation 1

x + y= 70

x= 70-y

Substitutes 70-y for x in equation 2

0.26(70-y) + 0.40y= 21

18.2-0.26y+0.40y= 21

18.2+0.14y= 21

0.14y= 21-18.2

0.14y= 2.8

Divide both sides by the coefficient of y which is 0.14

0.14y/0.14= 2.8/0.14

y= 20

Substitute 20 for y in equation 1

x + y= 70

x + 20= 70

x= 70-20

x = 50

Hence Dan and june should use 50 pounds of the cheaper kind in the mix

step-by-step explanation:

sorry i just needed points for a quiz plz for give me

delaney would like to make a 5lb nut mixture that is 60% peanuts and 40% almonds.

she has several pounds of peanuts and several pounds of a mixture that is 20% peanuts and 80% almonds.

let 'p' represent the number of pounds of peanuts needed to make the new mixture, and  

let 'm' represent the number of pounds of the 80% almond-20% peanut mixture.

:

p + m = 5

therefore

m = (5-p)

write an equation based on the percent of peanuts

1p + .2m = .6(5)

replace m with (5-p)

p + .2(5-p) = .6(5)

p + 1 - .2p = 3

p - .2p = 3 - 1

.8p = 2

p = 2/.8

p = 2.5 lbs of peanuts required

then

5 - 2.5 = 2.5 lbs of the peanut/almond mixture required

:

a) what is the system that models this situation?

p + m = 5

1p + .2m = .6(5)