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3 years ago

Pls someone answer me in correct way

Mathematics

2 answers:

Alex_Xolod [135]3 years ago

8 0

0/2=0
.........................

pogonyaev3 years ago

3 0

2nd line:
output y=0, ordered pair (0, 0)

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2 years ago

Find the equation of locus of a point which moves such that its distance from (0,2) is one third distance from (-2,3). ( I WILL

scoray [572]

Answer:

$8(x^2+y^2)-4x-30y+23=0$

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<u />

<u>Distance formula</u>

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

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Let B = (-2, 3)

If a point moves such that its distance from (0, 2) is one third distance from (-2, 3):

$PA=\dfrac{1}{3}PB$

Therefore, using the distance formula:

$\implies \sqrt{(x-0)^2+(y-2)^2}=\dfrac{1}{3}\sqrt{(x-(-2))^2+(y-3)^2}$

Square both sides:

$\implies x^2+(y-2)^2=\dfrac{1}{9}[(x+2)^2+(y-3)^2]$

$\implies x^2+y^2-4y+4=\dfrac{1}{9}(x^2+4x+4+y^2-6y+9)$

Multiply both sides by 9:

$\implies 9x^2+9y^2-36y+36=x^2+4x+4+y^2-6y+9$

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$\implies 8(x^2+y^2)-4x-30y+23=0$

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2 years ago

G-1/6=1/6. Solve for G

aleksklad [387]

G -1/6= 1/6

Move -1/6 to the other side

sign changes from -1/6 to 1/6

G-1/6+1/6=1/6+1/6

G = 2/6

Reducing: divide by 2 for 2/6

2/2= 1

2/6= 3

Answer : G= 2/6= 1/3

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3 years ago