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eimsori [14]
3 years ago
10

What is the value of 1/3 x^2 + 5.2y when x = 3 and y = 2? PLZ HELP!!!

Mathematics
1 answer:
olasank [31]3 years ago
6 0
The first step in solving this is to substitute x with 3 and y with 2, so that
\frac{1}{3} x^{2} +5.2y
becomes
\frac{1}{3}  (3)^{2} +5.2(2)
Then simply solve using the order of operations.
First exponents (there are no parentheses)
\frac{1}{3} 9 +5.2(2)
Then multiply and divide
3+10.4
And finally add and subtract
13.4
Therefore the value <span>of 1/3 x^2 + 5.2y when x=3 and y=2 is 13.4</span>
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  1. 6x +y = -6
  2. 6x -y = 8
  3. 5x +y = 13

Step-by-step explanation:

To rewrite these equations from point-slope form to standard form, you can do the following:

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__

1. y -6 = -6(x +2)

  y -6 = -6x -12 . . . . . eliminate parentheses

  6x +y -6 = -12 . . . . . add 6x

  6x +y = -6 . . . . . . . . add 6

__

2. y +2 = 6(x -1)

  y +2 = 6x -6

  -6x +y +2 = -6

  -6x +y = -8

  6x -y = 8 . . . . . . . . multiply by -1

__

3. y -3 = -5(x -2)

  y -3 = -5x +10

  5x +y -3 = 10

  5x +y = 13

_____

<em>Additional comment</em>

The "standard form" of a linear equation is ax+by=c for integers a, b, c. The leading coefficient (generally, 'a') should be positive, and all coefficients should be mutually prime (have no common factors). That is why we multiply by -1 in problem 2.

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