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timofeeve [1]
3 years ago
12

Stuck on this it's Algebra 2 12x^{3} -3x^{2} =0

Mathematics
1 answer:
olchik [2.2K]3 years ago
6 0

Answer:

Exact Form:

x = 0, 1/4

Decimal Form:

x = 0, 0.25

Step-by-step explanation:

<u>Step 1: Factor 3x^2 out of 12x^3 - 3x^2 </u>

Factor 3x^2 out of 12x^3:

<em> 3x^2 (4x) - 3x^2 = 0</em>

Factor 3x^2 out of -3x^2:

<em> 3x^2 (4x) + 3x^2 (-1)  = 0</em>

Factor 3x^2 out of -3x^2 <em>(4x) + 3x^2 (-1) </em>:

<em> 3x^2 (4x-1) = 0</em>

<em />

<u>Step 2: Divide each term by 3 and simpify</u>

<u></u>

divide each term in 3x^2 (4x-1) = 0 by 3.

3x^2 (4x-1) / 3 = 0 / 3

<em>simplify 3x^2 (4x-1) / 3.</em>

<em />

<em>Cancel the common factors.</em>

<u><em>3</em></u><em> </em>x^2 (4x -1) / <em><u>3</u></em> = 0 / 3

divide x^2 (4x-1) by 1.

x^2 (4x-1) / 3 = 0 / 3

Apply the Distributive Property

Reorder.

Rewrite using the commutative property of multiplication.

4x^2 x + x^2 · -1 = 0 / 3

Move -1  to the left of  x^2

4x^2 x -1 · x^2  = 0 / 3

Simplify each term

multiply x^2 by x^2 by adding the exponents.

Move x

4 (x · x^2) -1 · x^2= 0 / 3

Multiply x by x^2

Rase x to the power of 1.

4 (x^1 · x^2) -1 · x^2= 0 / 3

Use the power rule a^m a^n = a^m+n to combine exponents

4x^1+2 -1 · x^2= 0 / 3

Add 1 and 2.

4x^3 -1 · x^2= 0 / 3

Rewrite -1x^2  as -x^2.

4x^3 -x^2= 0 / 3

Divide 0 by 3

4x^3 -x^2= 0

<u>Step 3: Factor x^2 out of 4x^3 -x^2.</u>

Factor x^2 out of 4x^3

x^2 (4x) -x^2 = 0

Factor x^2 out of -x^2

x^2 (4x) x^2 · -1 = 0

Factor x^2 out of x^2 (4x) x^2 · -1

x^2 (4x -1) = 0

If any individual factor on the left side of the equation is equal to 0,  the entire expression will be equal to  0

x^2 = 0

4x -1 = 0

Set x^2 equal to 0 and solve for x

x = 0

Set 4x -1 equal to 0 and solve for x

x = 1/4

The final solution is all the values that make x^2 (4x-1) = 0 true

x= (0, 1/4)

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Below

Step-by-step explanation:

For a given shape to be a rhombus, it should satisfy these conditions:

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● The diagonals should be perpendicular.

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We will prove the conditions one by one.

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's prove that the diagonals are perpendicular:

To do that we will write express them as vectors

The two vectors are EG and DF.

The coordinates of the four points are:

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Now the coordinates of the vectors:

● EG (0-0,0-2c) => EG(0,-2c)

● DF ( a+b-(-a-b),c-c) => DF (2a+2b,0)

For the diagonals to be perpendicular the scalar product of EG and DF should be null.

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So the diagonals are perpendicular.

■■■■■■■■■■■■■■■■■■■■■■■■■■

Let's prove that the diagonals intercept each others at the midpoints.

The diagonals EG and DF should have the same midpoint.

● The midpoint of EG:

We can figure it out without calculations. Since G is located at (0,0) and E at (0,2c) then the distance between E and G is 2c.

Then the midpoint is located at (0,c)

● The midpoint of DF:

We will use the midpoint formula.

The coordinates of the two points are:

● F (a+b,c)

● D(-a-b,c)

Let M be the midpoint of DF

●M( (a+b-a-b,c+c)

● M (0,2c)

So EG and DF have the same midpoint.

■■■■■■■■■■■■■■■■■■■■■■■■■■

There is no need to prove the last condition, since the two above guarante it.

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Step-by-step explanation:

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