Simplify this expression

Mathematics

2 answers:

inessss [21]3 years ago

7 0

Answer:

3x^2+27y^2-18xy

Step-by-step explanation:

3(x-3y)^2

3 [(x-3y)(x-3y)]

3 [x^2-3xy-3xy+9y^2]

3 [x^2-6xy+9y^2]

open d bracket

3x^2-18xy+27y^2

rearrange

3x^2+27y^2-18xy

vodomira [7]3 years ago

3 0

Answer:

3x² - 18xy + 27y²

Step-by-step explanation:

= 3 ( x - 3y )²

= 3 ( x² - 2 * -3y * x + ( -3y)²)

= 3( x² - 6xy +9y²)

= 3x² - 18xy + 27y²

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den301095 [7]

Answer:

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Step-by-step explanation:

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8 0

3 years ago

Read 2 more answers

Lamar is writing a coordinate proof to show that a segment from the midpoint of the hypotenuse of a right triangle to the opposi

Zanzabum

1. N is a midpoint of the segment KL, then N has coordinates

$\left(\dfrac{x_K+x_L}{2},\dfrac{y_K+y_L}{2} \right) =\left(\dfrac{0+2a}{2},\dfrac{2b+0}{2} \right) =(a,b).$

2. To find the area of △KNM, the length of the base MK is 2b, and the length of the height is a. So an expression for the area of △KNM is

$A_{KMN}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2b\cdot a=ab.$

3. To find the area of △MNL, the length of the base ML is 2a and the length of the height is b. So an expression for the area of △MNL is

$A_{MNL}=\dfrac{1}{2}\cdot \text{base}\cdot \text{height}=\dfrac{1}{2}\cdot 2a\cdot b=ab.$

4. Comparing the expressions for the areas you have that the area $A_{KMN}$ is equal to the area $A_{MNL}$ . This means that the segment from the midpoint of the hypotenuse of a right triangle to the opposite vertex forms two triangles with equal areas.