Hi,

$18x^2+36y^2=648\\

\Rightarrow ( \dfrac{x}{6})^2+( \dfrac{y}{3* \sqrt{2} } )^2=1\\

a=6,\ b=3* \sqrt{2}\\

c= \sqrt{a^2-b^2} = \sqrt{36-18} =3* \sqrt{2}\\

$
Center is (0,0)
Foci are (-3√2,0) and (3√2,0)

7 0

2 years ago

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I NEED HELP FASTT!!! PLEASE

AysviL [449]

Answer:

Step-by-step explanation:

8x - 6 > 12 + 2x

Subtract 2x from each

8x -2x -6 > 12+2x-2x

6x -6 > 12

Add 6 to each side

6x -6+6 > 12+6

6x > 18

Divide by 6

6x/6 > 18/6

x > 3

The only value greater than 3 is 5

4 0

2 years ago

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What is (-7) x (-3) in multiplying integers

Roman55 [17]

.... Hope this will help....

6 0

2 years ago

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Simplify the ratio:<br> 18,750 : 216 =

aleksandr82 [10.1K]

The ratio to this problem is 750:216

5 0

2 years ago

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