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BabaBlast [244]
3 years ago
9

5Csqrt%7B%20%5Cleft%7C%20x%20%5Cright%7C%20%7D%20%5Cright%29%20%7D%5E%7B%202%20%7D%20%3D%201" id="TexFormula1" title=" \large \tt \: { x }^{ 2 } + { \left( y- \sqrt{ \left| x \right| } \right) }^{ 2 } = 1" alt=" \large \tt \: { x }^{ 2 } + { \left( y- \sqrt{ \left| x \right| } \right) }^{ 2 } = 1" align="absmiddle" class="latex-formula">
Solve for y. Attach a graph too.
Note :- The graph will come in the shape of a heart.

Only solve if you know it!​​
Mathematics
2 answers:
Gennadij [26K]3 years ago
8 0

Refer to the attachment

wel3 years ago
5 0

\huge \boxed{\mathbb{QUESTION} \downarrow}

  • \large \tt \: { x }^{ 2 } + { \left( y- \sqrt{ \left| x \right| } \right) }^{ 2 } = 1

\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}

{ x  }^{ 2  }  + {  \left( y- \sqrt{  \left| x  \right|    }    \right)    }^{ 2  }   =  1

Subtract x² from both sides of the equation.

\left(y-\sqrt{|x|}\right)^{2}+x^{2}-x^{2}=1-x^{2}

Subtracting x² from itself leaves 0.

\left(y-\sqrt{|x|}\right)^{2}=1-x^{2}

Take the square root of both sides of the equation.

y-\sqrt{|x|}=\sqrt{1-x^{2}}  \\ y-\sqrt{|x|}=-\sqrt{1-x^{2}}

Subtract − √∣x∣ from both sides of the equation.

y-\sqrt{|x|}-\left(-\sqrt{|x|}\right)=\sqrt{1-x^{2}}-\left(-\sqrt{|x|}\right)  \\ y-\sqrt{|x|}-\left(-\sqrt{|x| } \right)=-\sqrt{1-x^{2}}-\left(-\sqrt{|x|}\right)

Subtracting − √∣x∣ from itself leaves 0.

y=\sqrt{1-x^{2}}-\left(-\sqrt{|x|}\right) \\  y=-\sqrt{1-x^{2}}-\left(-\sqrt{|x|}\right)

Subtract − √∣x∣from √1- x².

\underline{\underline{ \sf \: y=\sqrt{1-x^{2}}+\sqrt{|x|} }}

Subtract − √∣x∣from - √1- x².

\underline{\underline{ \sf \: y= - \sqrt{1-x^{2}}+\sqrt{|x|} }}

The equation is now solved.

\large \boxed{ \boxed{ \bf \: y=\sqrt{1-x^{2}}+\sqrt{|x|} }}\\   \\   \large\boxed {\boxed{ \bf \: y=-\sqrt{1-x^{2}}+\sqrt{|x|} }}

_________________________________

  • Refer to the attached image for the graph.

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Ulleksa [173]
First one:

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when adding the like terms
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placehloder letter (x adds with x and y adds with y and so on)

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powers: m^1 and M^4
placeholders: all m


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second one:
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Triangle S R Q is shown. Angle S R Q is a right angle. An altitude is drawn from point R to point T on side S Q to form a right
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Step-by-step explanation:

From the figure attached,

ΔSRQ is right triangle.

m∠R = 90°

An altitude has been constructed from point T to side SQ.

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By applying geometric mean theorem in triangle SRQ,

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x² = 16 × 9

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x = √144

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8_murik_8 [283]

Answer:

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