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ycow [4]
3 years ago
11

How many 68 times fit in 345

Mathematics
2 answers:
sleet_krkn [62]3 years ago
8 0

5 time because you wanna multiply 68 times a number that gets you the closest to 345 which is 68 x 5 = 340 and 68 x 6 = 408 which is too high.

Arisa [49]3 years ago
7 0
About 5 times.........
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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

Determine 2a - 3b + 4c

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

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

((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

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

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

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

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

Express 0 as 0 + 0 + 0

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

By comparison

(a - 1)^2 = 0

(b + 1)^2 = 0

(c  + 1)^2 = 0

Solving for (a - 1)^2 = 0

Take square root of both sides

a - 1 = 0

Add 1 to both sides

a - 1 + 1 = 0 + 1

a = 1

Solving for (b + 1)^2 = 0

Take square root of both sides

b + 1 = 0

Subtract 1 from both sides

b + 1 - 1 = 0 - 1

b = -1

Solving for (c  + 1)^2 = 0

Take square root of both sides

c + 1 = 0

Subtract 1 from both sides

c + 1 - 1 = 0 - 1

c = -1

Substitute the values of a, b and c in 2a - 3b + 4c

2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)

2a - 3b + 4c = 2 +3  -4

2a - 3b + 4c = 1

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