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Digiron [165]
3 years ago
8

During the first week of school Andrew purchased two pounds of Jellies and two pounds of Bonbons for $38. The second week of sch

ool he purchased two pounds of Jellies and four pounds of Chocolates for $74 and the third week he purchased two pounds of Jellies, two pounds of Bonbons and one pound of Chocolates for $53. What is the cost of one pound of Chocolates?
Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
7 0

Answer:

j = $7

b = $12

c = $15

Step-by-step explanation:

Let

j = cost of one pound of jelly

b = cost of one pound of Bonbon

c = cost of one pound of chocolate

First week:

two pounds of Jellies and two pounds of Bonbons for $38.

2j + 2b = 38

Second week:

two pounds of Jellies and four pounds of Chocolates for $74

2j + 4c = 74

Third week:

two pounds of Jellies, two pounds of Bonbons and one pound of Chocolates for $53.

2j + 2b + c = 53

Recall,

2j + 2b = 38

So,

38 + c = 53

c = 53 - 38

= 15

c = $15

Substitute c = 15 into

2j + 4c = 74

2j + 4(15) = 74

2j + 60 = 74

2j = 74 - 60

2j = 14

j = 14/2

= 7

j = $7

Substitute j = 7 into

2j + 2b = 38

2(7) + 2b = 38

14 + 2b = 38

2b = 38 - 14

2b = 24

b = 24/2

= 12

b = $12

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Answer:

Step-by-step explanation:

If x - 1 is a factor of the cubic, then 1 should result in 0 when you put it in the cubic for x. Does it?

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f(1) = 2 - 15 + 22 + 15

f(1)= 24.

No the result is not zero. So x - 1 is not a factor.

However there must be something that makes this cubic go to zero. The easiest way to find it is to graph it.

The numbers that do make this zero are -0.5, 3, 5

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Does anyone get this​
ivolga24 [154]

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X would be gallons and y is cost.

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Answer:

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