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PilotLPTM [1.2K]
3 years ago
9

Please help me with this, easy points.

Mathematics
1 answer:
m_a_m_a [10]3 years ago
4 0

Elimination is where you aline two equations with the varibles in the same place and the second equation substracts the first equation. Important to know that in the second equation all varibles and constant or y - intercepts are negative. Now with this just start with the first two equations, 5x + 2y -3z = 10 and 2x - 2y + 4z = 6.

5x + 2y - 3z = 10

- 2x + 2y - 4z = -6

3x + 4y - 7z = 4

Do the same thing with the last equation.

3x + 4y - 7z = 4

-x + y - 2z = -3

2x + 5y -9z = 1

Please tell me if I got it wrong, but from what I concluded there's no solution.

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Use a protractor to do the problem, but try finding th answer yourself
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3 years ago
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Can anyone factor this?? I'm a little lost tysm!!!!<br> 5x^2+12x+4 5x 2 +12x+4
Sergio039 [100]

For a polynomial of the form ax^2+bx+c rewrite the middle term as a sum of two terms whose product is a⋅c=5⋅4=20 and whose sum is b=12.

<u>Factor 12 out of 12x.</u>

5x^2+12(x)+4

<u>Rewrite 12 as 2 plus 10</u>

5x^2+(2+10)x+4

Apply the distributive property.

5x^2+2x+10x+4

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

(5x^2+2x)+10x+4

Factor out the greatest common factor (GCF) from each group.

x(5x+2)+2(5x+2)

Factor the polynomial by factoring out the greatest common factor, 5x+25x+2.

(5x+2)(x+2)

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2 years ago
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Ede4ka [16]
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Kimmy bought a 5 kilogram can of peanuts for $4.50. What is the unit price?
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3 years ago
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When the sum of \, 528 \, and three times a positive number is subtracted from the square of the number, the result is \, 120. F
aleksandr82 [10.1K]

Let x be the unknown number. So, three times that number means 3x, and the square of the number is x^2

We have to sum 528 and three times the number, so we have 528+3x

Then, we have to subtract this number from x^2, so we have

x^2-(3x+528)

The result is 120, so the equation is

x^2 - 3x - 528 = 120 \iff x^2 - 3x - 648 = 0

This is a quadratic equation, i.e. an equation like ax^2+bx+c=0. These equation can be solved - assuming they have a solution - with the following formula

x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

If you plug the values from your equation, you have

x_{1,2} = \dfrac{3\pm\sqrt{9-4\cdot(-648)}}{2} = \dfrac{3\pm\sqrt{9+2592}}{2} = \dfrac{3\pm\sqrt{2601}}{2} = \dfrac{3\pm51}{2}

So, the two solutions would be

x = \dfrac{3+51}{2} = \dfrac{54}{2} = 27

x = \dfrac{3-51}{2} = \dfrac{-48}{2} = -24

But we know that x is positive, so we only accept the solution x = 27

6 0
3 years ago
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