3 years ago

X2+Y2=25

Mathematics

1 answer:

Fofino [41]3 years ago

8 0

$x^{2} + y^{2} =25$ is equation of circle which have radius equal 5 and center in (0,0) point

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Consider a circle whose equation is x2 + y2 + 4x - 6y - 36 = 0. Which statements are true? Check all that apply.

kherson [118]

Answer: b & c

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3 years ago

Read 2 more answers

Can someone help me with at least one problem and tell me how you got the answer

Vesna [10]

S=16t^2
if t=3
So you would plug it in
16(3)^2
Then you multipy
16*3^2
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3 years ago

If x^2+1/x^2=3 find the value of x^2/(x^2+1)^2 Express answer as a common fraction. Thanks!!

timama [110]

$x^2+\dfrac1{x^2}=3\implies x^4+1=3x^2\implies x^4-3x^2+1=0$

By the quadratic formula,

$x^2=\dfrac{3\pm\sqrt5}2\implies x^2+1=\dfrac{5\pm\sqrt5}2$

Then

$(x^2+1)^2=\dfrac{25\pm10\sqrt5+5}4=\dfrac{15\pm5\sqrt5}2$

$\implies\dfrac{x^2}{(x^2+1)^2}=\dfrac{\frac{3\pm\sqrt5}2}{\frac{15\pm5\sqrt5}2}=\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}$

Multiply numerator and denominator by the denominator's conjugate:

$\dfrac{3\pm\sqrt5}{15\pm5\sqrt5}\cdot\dfrac{15\mp5\sqrt5}{15\mp5\sqrt5}=\dfrac{45\pm15\sqrt5\mp15\sqrt5-25}{15^2-(5\sqrt5)^2}=\dfrac{20}{100}=\dfrac15$