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3 years ago

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How do you simplify this?x²y+xy² / y²+2/5 × xy

1 answer:

polet [3.4K]3 years ago

3 0

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

How do you simplify this?
x²y+xy² / y²+2/5 × xy

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\sf\frac{ { x }^{ 2 } y+x { y }^{ 2 } }{ { y }^{ 2 } + \frac{ 2 }{ 5 } \times xy } \\$

Factor the expressions that are not already factored.

_____

<u>How </u><u>to</u><u> factorise</u><u> </u><u>:</u><u>-</u>

<u>NUMERATOR</u> $\downarrow$

$\sf \: x ^ { 2 } y + x y ^ { 2 }$

Factor out xy.

$\sf \: xy\left(x+y\right)$

<u>DENOMINATOR</u> $\downarrow$

$\sf{ y }^{ 2 } + \frac{ 2 }{ 5 } \times xy \\$

Factor out 1/5.

$\sf \: {\frac{1}{5}y\left(2x+5y\right)} \\$

_____

Continuing...

$\sf\frac{xy\left(x+y\right)}{\frac{1}{5}y\left(2x+5y\right)} \\$

Cancel out y in both the numerator and denominator.

$\sf\frac{x\left(x+y\right)}{\frac{1}{5}\left(2x+5y\right)} \\$

Expand the expression.

$\sf\frac{x^{2}+xy}{\frac{2}{5}x+y} \\$

This can further simplified to as $\downarrow$

$= \boxed{\boxed{\bf\frac{5x\left(x +y\right)}{2x+5y}}}$

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The corners of a meadow are shown on a coordinate grid. Ethan wants to fence the meadow. What length of fencing is required?

Nuetrik [128]

Answer:

34.6 units

Step-by-step explanation:

The lenght of fencing required is the total distance between point A to B, B to C, C to D, and D to A. That is the distance between all 4 corners of the meadow.

The coordinates of the corners of the meadow is shown on a coordinate plane in the attachment. (See attachment below).

Let's use the distance formula to calculate the distance between the 4 corners of the meadow using their coordinates as follows:

Distance between point A(-6, 2) and point B(2, 6):

$AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$A(-6, 2)) = (x_1, y_1)$

$B(2, 6) = (x_2, y_2)$

$AB = \sqrt{(2 - (-6))^2 + (6 - 2)^2}$

$AB = \sqrt{(8)^2 + (4)^2}$

$AB = \sqrt{64 + 16} = \sqrt{80}$

$AB = 8.9$ (nearest tenth)

Distance between B(2, 6) and C(7, 1):

$BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$B(2, 6) = (x_1, y_1)$

$C(7, 1) = (x_2, y_2)$

$BC = \sqrt{(7 - 2)^2 + (1 - 6)^2}$

$BC = \sqrt{(5)^2 + (-5)^2}$

$BC = \sqrt{25 + 25} = \sqrt{50}$

$BC = 7.1$ (nearest tenth)

Distance between C(7, 1) and D(3, -5):

$CD = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$C(7, 1) = (x_1, y_1)$

$D(3, -5) = (x_2, y_2)$

$CD = \sqrt{(3 - 7)^2 + (-5 - 1)^2}$

$CD = \sqrt{(-4)^2 + (-6)^2}$

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$CD = 7.2$ (nearest tenth)

Distance between D(3, -5) and A(-6, 2):

$DA = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let,

$D(3, -5) = (x_1, y_1)$

$A(-6, 2) = (x_2, y_2)$

$DA = \sqrt{(-6 - 3)^2 + (2 - (-5))^2}$

$DA = \sqrt{(-9)^2 + (7)^2}$

$DA = \sqrt{81 + 49} = \sqrt{130}$

$DA = 11.4$ (nearest tenth)

Length of fencing required = 8.9 + 7.1 + 7.2 + 11.4 = 34.6 units

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