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neonofarm [45]
3 years ago
15

A store is having a sale on jelly beans and trail mix. For 3 pounds of jelly beans and 2 pounds of trail mix, the total cost is

$11. For 5 pounds of jelly beans and 6 pounds of trail mix, the total cost is $23 . Find the cost for each pound of jelly beans and each pound of trail mix.
Mathematics
1 answer:
Nata [24]3 years ago
6 0

the cost for each of jelly beans and each pound of trail mix is $2.5 and $1.75

<u>Step-by-step explanation:</u>

Given A store is having a sale on jelly beans and trail mix. For 3 pounds of jelly beans and 2 pounds of trail mix, the total cost is $11. For 5 pounds of jelly beans and 6 pounds of trail mix, the total cost is $23 . We have to find the cost for each pound of trail mix and each pound of jelly beans.

Let the cost of each pound of trail mix is $y.

and  the cost of each pound of jelly beans is $x.

According to question,

For   3 pounds of jelly beans and 2 pounds of trail mix, the total cost is $11.

⇒ 3x+2y=11  →  (1)

For 5 pounds of jelly beans and 6 pounds of trail mix, the total cost is $23

⇒ 5x+6y=23 →  (2)

Solving (1) and (2), we get

3(1 equation)-(2 equation)=0

⇒3(3x+2y)-5x-6y=3(11)-23

⇒9x+6y-5x-6y=10

hence,

⇒4x=10

⇒x=2.5

Putting x=2.5 in  3x+2y=11  we get ;

⇒  3x+2y=11

⇒  3(2.5)+2y=11

⇒  2y=11-7.5=3.5

⇒  y=1.75

Hence, the cost for each of jelly beans and each pound of trail mix is $2.5 and $1.75 .

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Answer:

y = 30x +50 --- Sam

y = 20x +80 --- Tim

Step-by-step explanation:

Given

Sam                         Tim

х  --- f(x) ---------------  g(x)

1  --- 80   --------------- 100

2  --- 110  --------------- 120

3  --- 140 --------------- 140

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Required

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First, we need to take any corresponding values of x and y

(x_1,y_1) = (1,80)

(x_2,y_2) = (4,170)

Next, we calculate the slope (m)

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m = \frac{90}{3}

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Next, we calculate the line equation using:

y - y_1=m(x-x_1)

y - 80 = 30(x - 1)

y - 80 = 30x - 30

Make y the subject

y = 30x - 30 + 80

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First, we need to take any corresponding values of x and y

(x_1,y_1) = (1,100)

(x_2,y_2) = (4,160)

Next, we calculate the slope (m)

m = \frac{y_2 - y_1}{x_2 - x_1}

m = \frac{160 - 100}{4 - 1}

m = \frac{60}{3}

m = 20

Next, we calculate the line equation using:

y - y_1=m(x-x_1)

y - 100 = 20(x - 1)

y - 100 = 20x - 20

Make y the subject

y = 20x - 20+100

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6-4(2x-3)+10x

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