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3 years ago

What is the point of intersection when the system of equations below is graphed on the coordinate plane?

Mathematics

1 answer:

WARRIOR [948]3 years ago

8 0

Neither of the options are correct for the given equations.

Step-by-step explanation:

-x+y=4...................................................1

16x+y=-3...............................................2

Subtract equation 1 and 2

-x+y- 16x -y = 4+3

⇒ -17x = 7

⇒x= $\frac{7}{-17\\}$

⇒x= -0.4117

so,

16×-0.4417 + y = -3

⇒y= -3+6.5872

⇒y= 3.5872

So the point of intersection is (-0.4117, 3.5872)

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Find the equation of the circle with a diameter whose end points are (-1,-2) and (-3,2)

Sergeeva-Olga [200]

Answer:

The equation of the circle with a diameter whose end points are (-1,-2) and (-3,2) is

$x^{2} +y^{2}+4x-1=0$

Step-by-step explanation:

Explanation:-

Step 1:-

The equation of the circle having center and radius is

$(x-h)^2+(y-k)^2=r^2$

here center is (h,k) and radius is r

Given diameter whose end points are (-1,-2) and (-3,2)

The diameter of the circle is passing through the center of the circle

so center of the circle = midpoint of two end points

$(\frac{-1 +(-3) }{2} ,\frac{-2+2 }{2} )$

$(-2,0)$

therefore center (h,k) = (-2,0)

Step 2:-

we have to find the radius of the circle

the radius of the circle = the distance from center to the one end point

i.e., C P = r

Given one end point is P(-3,2) and center C(-2,0)

The distance formula of two points are

$\sqrt{(x_{2}-x_{1} ) ^{2}+ (y_{2}-y_{1} ) ^{2}}$

$r=\sqrt{{(-3)-(-1) ) ^{2}+ (2-(-2)) ^{2}}$

$r=\sqrt{5}$

Step 3:-

center (h,k) = (-2,0) and

radius $r=\sqrt{5}$

The standard form of circle equation

$(x-h)^2+(y-k)^2=r^2$

$(x-(-2))^2+(y-0)^2=\sqrt{5} ^2$

on simplification is

$x^{2} +y^{2}+4 x-1=0$

5 0

3 years ago

X = y−4, 2x−5y = 3

dalvyx [7]

X=y-4 Let's make this eqn 1 and

2x-5y=3 this eqn 2..

Two things are possible here...
Its either you substitute for x in equation 2... Or you substitute for y...

x=y-4 put this in the equation 2..

2x-5y=3.. Replace the x with (y-4)

2(y-4)-5y = 3...
Since this is In the option no point in substituting for y...

Hope this helped...?

5 0

3 years ago

Product of 2a+3b-c and 2a+3b-c

goldenfox [79]

Answer:

Product (multiplication) of 2a+3b-c and 2a+3b-c is −2ac+2a+6b−c.

Explanation:

2a+3b−c(2)a+3b−c

= 2a+3b+−2ac+3b+−c

Combine Like Terms:

= 2a+3b+−2ac+3b+−c

= (−2ac)+(2a)+(3b+3b)+(−c)

= −2ac+2a+6b+−c

6 0

3 years ago

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If A parking garage charges three dollars for every hour how much would the cost be to park for t hours

SSSSS [86.1K]

Answer:

hope this helps

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Step-by-step explanation:

3 0

2 years ago

Need help with this asap

dem82 [27]

Answer:

I cant lie I don't know this answer

4 0

2 years ago

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