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3 years ago

What is the polar form of the equation? x^2+(y+5)^2=25

Mathematics

1 answer:

Blizzard [7]3 years ago

6 0

Answer:

x=√-y^2-10y

Step-by-step explanation:

step 1:add y^-2 to both sides

x^2+(y+5)^2=25

x^2+y^2+10y+25+-y^2=25+-y^2

x^2+10y+25=-y^2+25

Step 2:add -10y to both sides

x^2+10y+25-10y=-y^2+25+-10y

x^2+25=-y^2-10y+25

step 3:add -25 to both sides

x^2+25-25=y^2-10y+25-25

x^2=-y^2-10y

step 4:take square root

x=√−y^2−10y

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3 years ago

Which equation represents the value of x? x=100−y2−−−−−−−√ x=10+y2 x=10−y x=y2+100−−−−−−−√ Right triangle A B C with angle B as

Nostrana [21]

Answer:

Therefore the required value of x,

$x=\sqrt{(100-y^{2})}$

Step-by-step explanation:

Given:

ΔBC is a Right Angle Triangle at ∠ B = 90°

As ∠ B = 90° , AC will be the Hypotenuse

AC = 10 = Hypotenuse

BC = y = Longer leg ( say )

AB = x = Shorter leg ( say )

To Find :

x = ?

Solution:

In Right Angle Triangle Δ ABC , By Pythagoras Theorem we get

$(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}$

Substituting the given values we get

$10^{2}= x^{2}+y^{2} \\\\x^{2}=100-y^{2} \\\\\textrm{square rooting on both the side we get}\\\\x=\sqrt{(100-y^{2})}\\\therefore x=\sqrt{(100-y^{2})}\ \textrm{ which is the required value of x}$