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marin [14]
3 years ago
8

In the system shown below, what are the coordinates of the solution that lies in quadrant I?

Mathematics
2 answers:
dimulka [17.4K]3 years ago
6 0

Answer:

(4,3)

Step-by-step explanation:

solve the simultaneous equations

x²+y²=25.........................(i)

x-y²= -5.............................(ii) ⇒make y² the subject of the equation;

y²=x+5.................................(iii)

Substitute equation (iii) in (i)

x² + x+5 =25

x²+x=25-5

x²+x-20=0..........solve for the quadratic equation

x(x-4)+5(x-4)=0

(x-4)(x+5)=0

x-4=0

x=  4 or

x+5=0

x=  -5

finding value of y

if y²=x+5 then;

when x=4, y²=4+5=9⇒y=√9 = ±3...................y= ±3

when x= -5  , y²= -5+ 5=0............... y=0

Coordinates = (4,3) and (-5,0)

Coordinates that lie in the 1st quadrant is (4,3)

Arisa [49]3 years ago
4 0

Answer:

(4,3)

Step-by-step explanation:

Given

x^2+y^2=25

x-y^2=-5

In order to solve the equations, from equation 2 we get

-y^2= -5-x

y^2=5+x

Putting the value of y^2 in equation 1

x^2+5+x=25

x^2+5-25+x=0

x^2+x-20=0

x^2+5x-4x-20= 0

x(x+5)-4(x+5)=0

(x+5)(x-4)=0

So

x+5=0  x-4=0

x=-5   x=4

Now for x=-5

x^2+y^2=25

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

25+y^2=25

y^2=25-25

y^2=0

so Y=0  

And for x = 4

x^2+y^2=25

(4)^2+y^2=25

16+y^2=25

y^2=25-16

y^2=9

y= ±3

So the solution to the system of equations is

(-5,0) , (4,3), (4,-3)

The only solution that belongs to first quadrant is (4,3)

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<em>Answer:</em>

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<em>Step-by-step explanation:</em>

<em>Rewrite the equation as </em>\sqrt{x} 6a+r/5r<em>  = m</em>

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<em>Then, you simplify each of the equation. </em>

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<em>Remove any parentheses if needed.</em>

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<em>Multiply each term by r and simplify."</em>

<em>Multiply both sides of the equation by  5.</em>

<em>6a+r= m^2r⋅(5)</em>

<em>Remove parentheses.</em>

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<em>6a+r=5m^2)r</em>

<em>Subtract  5m^2)r  from both sides of the equation.</em>

<em>6a+r-5m^2)r=0</em>

<em>Subtract  6a  from both sides of the equation.</em>

<em>r-5m^2)r=-6a</em>

<em>Factor  r out of  r-5m^2)r  </em>

<em>r(1-5m^2)=-6a</em>

Divide each term by 1-5m^2 and simplify.

r = - \frac{6a}{1-5m^2}

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