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BartSMP [9]
2 years ago
13

Corinne bought 5 bags of chips and 4 jars of dipping sauce for $21.82. At the same prices, Ginger bought 4 bags of chips and 3 j

ars of dipping sauce for $16.86. What is the price of one jar of dipping sauce?

Mathematics
1 answer:
mr_godi [17]2 years ago
7 0

Answer:

The price of one jar of dipping sauce is $2.98

Step-by-step explanation:

Let

x ----> the price of one bag of chip

y ----> the price of one jar of dipping sauce

we know that

<em>Corinne</em>

5x+4y=21.82 ----> equation A

<em>Ginger</em>

4x+3y=16.86 ----> equation B

Solve the system of equations by graphing

Remember that the solution is the intersection point both graphs

using a graphing tool

The solution is the point (1.98,2.98)

see the attached figure

therefore

The price of one bag of chip is $1.98

The price of one jar of dipping sauce is $2.98

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.
Verdich [7]

Answer:

2a - 3b + 4c = 1

Step-by-step explanation:

Given

a^2 + b^2 + c^2 = 2(a - b - c) - 3

Required

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a^2 + b^2 + c^2 = 2(a - b - c) - 3

Open bracket

a^2 + b^2 + c^2 = 2a - 2b - 2c - 3

Equate the equation to 0

a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0

Express 3 as 1 + 1 + 1

a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0

Collect like terms

a^2 - 2a + 1 + b^2 + 2b + 1 + c^2  + 2c + 1 = 0

Group each terms

(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

Factorize (starting with the first bracket)

(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

((a - 1)^2) + (b^2 + 2b + 1) + (c^2  + 2c + 1) = 0

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((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)(b + 1)) + (c^2  + 2c + 1) = 0

((a - 1)^2) + ((b + 1)^2) + (c^2  + 2c + 1) = 0

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Express 0 as 0 + 0 + 0

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