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

c: y > - 1/6x + 1

step-by-step explanation:

dashed line - meaning this < or > also not equal

solid line - meaning this < or > also equal

your answer is c because y > means up & it's shaded up, but y < means down

hope this ! : )

r = 78539458390583095803458309458039485093485 or just 3

step-by-step explanation: