-2
Equations: 3-x-5x2-5x = x+z
Rearrange into a quadratic equation: 5x2+2x+(-3+z) = 0
Use the discriminant to find the value of z:-
22-4*5*(-3+z) = 0
64-20z = 0
-20z = -64
z = 3.2
2
No, not if the y is squared. When graphed the equation will not form a straight line.
no
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
k = 0.1
2
No, not if the y is squared. When graphed the equation will not form a straight line.
no
equation 1: y = x-4 => y2 = x2-8x+16 when both sides are squared equation 2: x2+y2 = 8 Substitute equation 1 into equation 2: x2+x2-8x+16 = 8 => 2x2-8x+8 = 0 If the discriminant of the above quadratic equation is zero then this is proof that the line is tangent to the curve: The discriminant: b2-4ac = (-8)2-4*2*8 = 0 Therefore the discriminant is equal to zero thus proving that the line is tangent to the curve.
Circle equation: x^2 +y^2 -8x +4y = 30 Tangent line equation: y = x+4 Centre of circle: (4, -2) Slope of radius: -1 Radius equation: y--2 = -1(x-4) => y = -x+2 Note that this proves that tangent of a circle is always at right angles to its radius
(2, -2)
k = 0.1
In trig, the secant squared divided by the tangent equals the hypotenuse squared divided by the product of the opposite and adjacent sides of the triangle.Details: secant = hypotenuse/adjacent (H/A) and tangent = opposite/adjacent (A/O);Then secant2/tangent = (H2/A2)/(O/A) = H2/A2 x A/O = H2/AO.
Circle equation: x^2 +y^2 -8x -16y -209 = 0 Completing the squares: (x-4)^2 +(y-8)^2 = 289 Centre of circle: (4, 8) Radius of circle 17 Slope of radius: 0 Perpendicular tangent slope: 0 Tangent point of contact: (21, 8) Tangent equation: x = 21 passing through (21, 0)
The general form of a line tangent to a circle is:y=mx+a(1+m2 )1/2. where "a" is the radius of the circle. Here circle is x2 + y2=4, so radius=a=2. nowc2=a2(1+m2)=8 (given)or 8=4(1+m2)2=1+m2 orm2=1 orm=1. so equation becomesy=mx+c ory=x+cImproved Answer:Equation 1: y = x+square root of 8 => y2 = x2+square root of 32x+8 when both sides are squared.Equation 2: y2 = 4-x2By definition:x2+the square root of 32x+8 = 4-x2 => 2x2+the square root of 32x+4 = 0If the discriminant b2-4ac of the above quadratic equation is equal to zero then this is proof that the straight line is tangent to the curve:b2-4ac = the square root of 322-4*2*4 = 0Therefore the straight line is a tangent to the curve because the discriminant of the quadratic equation equals zero.
x2 + y2 = 49
Use tangent. Your equation will be tan(slope of hypotenuse) = opposite side / adjacent side. it's easier if you just do A squared plus b squared equals c squared. Then subtitute the numbers gived in.