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elena-s [515]
3 years ago
8

PLZ HELP ASAP Geometry help: 2x^2 + 2y^2 - 8x +10y +2=0

Mathematics
1 answer:
Alexxx [7]3 years ago
6 0
Ok so this is conic sectuion
first group x's with x's and y's with y's
then complete the squra with x's and y's


2x^2-8x+2y^2+10y+2=0
2(x^2-4x)+2(y^2+5y)+2=0
take 1/2 of linear coeficient and square
-4/2=-2, (-2)^2=4
5/2=2.5, 2.5^2=6.25
add that and negative inside

2(x^2-4x+4-4)+2(y^2+5y+6.25-6.25)+2=0
factor perfect squares
2((x-2)^2-4)+2((y+2.5)^2-6.25)+2=0
distribute
2(x-2)^2-8+2(y+2.5)^2-12.5+2=0
2(x-2)^2+2(y+2.5)^2-18.5=0
add 18.5 both sides
2(x-2)^2+2(y+2.5)^2=18.5
divide both sides by 2
(x-2)^2+(y+2.5)^2=9.25

that is a circle center (2,-2.5) with radius √9.25

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Novosadov [1.4K]
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8 0
3 years ago
the combined age of april and laura is 23 years. laura's age is two years more than half of april's age. what is laura's age
Kipish [7]

Laura's age is 9 years.

Solution:

Let x be the age of April.

Laura's age = 2 years more than half of April's age

<u>Convert statement into algebraic expression:</u>

Half of April's age = \frac{x}{2}

2 years more than half of April's age = \frac{x}{2}+2

Combined age of April and Laura = 23

⇒ April's age + Laura's age = 23

$\Rightarrow \ \ x+(\frac{x}{2}+2) =23

$\Rightarrow \ \ x+\frac{x}{2}+2 =23

$\Rightarrow \ \ \frac{x}{1}+\frac{x}{2}+\frac{2}{1} =23

To add the fractions make the denominators same.

Multiplying 2 on both numerator and denominator of unlike terms, we get

$\Rightarrow \ \ \frac{x\times2}{1\times2}+\frac{x}{2}+\frac{2\times2}{1\times2} =23

$\Rightarrow \ \ \frac{2x}{2}+\frac{x}{2}+\frac{4}{2} =23

Denominators are same, now add the fractions.

$\Rightarrow \ \ \frac{2x+x+4}{2} =23

Do cross multiplication.

$\Rightarrow \ \ 2x+x+4=23\times2

$\Rightarrow \ \ 3x=46-4

$\Rightarrow \ \ 3x=42

$\Rightarrow \ \ x=14

Aprils's age = 14 years

Laura's age = \frac{14}{2}+2

                    = 7 + 2

Laura's age = 9

Hence Laura's age is 9 years.

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4 years ago
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bixtya [17]

Answer:

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Step-by-step explanation:

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=(2⋅ℎ⋅ℎ)+(2⋅ℎ⋅ℎℎ)+(2⋅ℎ⋅ℎℎ)

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