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What is the answer for x x -6x-8 equals -56?
x² - 6x -8 = -56 <=> x² -6x + 48 = 0 We now calculate the discriminant (which equals b²-4ac for an equation of the form ax² + bx + c-: D = b²-4ac = (-6)² - 4*1*48 = 36 - 192 = -156 I don't know what kind of course in calculus you are taking, but if you only want the real answers to this equation, you can stop here because the discriminant is negative, meaning there are no real solutions. However, there are complex solutions to this equation The complex roots of D are sqrt(156)*i=12,5*i and -sqrt(156)*i=-12,5*i There are two solutions to a quadratic equation, namely: x1 = (-b + sqrt(D))/(2*a) and x2 = (-b - sqrt(D))/(2*a) so the two solutions we find for this equation are: x1 = (-b + sqrt(D))/(2*a) = (6+12,5i)/2 = 3+6,25i and x2 = (-b - sqrt(D))/(2*a) = (6-12,5i)/2 = 3-6,25i x1 and x2 are complex solutions to this quadratic equation.
Write the equation in slope-intercept form of the line that has a slope of 2 and contains the point 3 7?
Use the slope intercept form here, Y - Y1 = m(X - X1) m = 2 Y1 and X1 = (3, 7) (Y - 7) = 2(X - 3) Y - 7 = 2X - 6 Y = 2X + 1 ----------------the equation of the line
What equation has a graph perpendicular to y equals 2x-3?
5
What is the equation of the line that goes through -1 3 and 5 -9?
(-1,3),(5,-9) (y2-y1)/(x2-x1)= (-9-3)/(5--1)= (-12/6)= the slope is -2 m(x-x1)=y-y1 -2(x-5)=y--9 -2x+10=y+9 y=-2x+1
Is the equation x plus y equals 4 linear or not?
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
X2 plus 2x - 15 equals 0?
x²+2x-15=0 x1=-2/2 - Square root of ((2/2)²+15) x1=-1-4 x1=-5 x2=-2/2 + Square root of ((2/2)²+15) x2=-1+4 x2=3
What is the answer for x x -6x-8 equals -56?
x² - 6x -8 = -56 <=> x² -6x + 48 = 0 We now calculate the discriminant (which equals b²-4ac for an equation of the form ax² + bx + c-: D = b²-4ac = (-6)² - 4*1*48 = 36 - 192 = -156 I don't know what kind of course in calculus you are taking, but if you only want the real answers to this equation, you can stop here because the discriminant is negative, meaning there are no real solutions. However, there are complex solutions to this equation The complex roots of D are sqrt(156)*i=12,5*i and -sqrt(156)*i=-12,5*i There are two solutions to a quadratic equation, namely: x1 = (-b + sqrt(D))/(2*a) and x2 = (-b - sqrt(D))/(2*a) so the two solutions we find for this equation are: x1 = (-b + sqrt(D))/(2*a) = (6+12,5i)/2 = 3+6,25i and x2 = (-b - sqrt(D))/(2*a) = (6-12,5i)/2 = 3-6,25i x1 and x2 are complex solutions to this quadratic equation.
What is the equation of the straight line passing through the point 4 -2 and perpendicular to the line whose equation is 2x -y -5 equals 0?
2x-y -5 = 0 => y = 2x -5 The perpendicular slope is the negative reciprocal of 2 which is -1/2 So using the formula of y -y1 = m(x -x1) gives the straight line equation:- y - -2 = -1/2(x -4) y +2 = -1/2x +2 y = -1/2x +2 -2 y = -1/2x which can be expressed in the form of: x +2y = 0 So the straight line equation is: x +2y = 0
Give an example of an equation that is an identity?
An identity is an equation that is always true, for any value of the variable or variables. Here are some examples: x + x = 2x a + b = b + a x1 = x
If f is a differentiable function such that f of 9 equals 2 and f' of 9 equals -2 then what is the linearization?
Use the point-slope form: y - y1 = m(x - x1). From the givens, y1 = 2, x1 = 9, and m = -2. Thus, y - 2 = -2(x - 9) = -2x + 18, or y = -2x + 20.
Write the equation in slope-intercept form of the line that has a slope of 2 and contains the point 3 7?
Use the slope intercept form here, Y - Y1 = m(X - X1) m = 2 Y1 and X1 = (3, 7) (Y - 7) = 2(X - 3) Y - 7 = 2X - 6 Y = 2X + 1 ----------------the equation of the line
How do you factor 2x 2-2x-2 using complex numbers?
To factor this, write the equation 2x2 - 2x - 2 = 0, and solve with the quadratic equation. You may get two numbers, which we may call "x1" and "x2"; in this case, the factorization is (x - x1)(x - x2).
X plus one over x squared plus two?
it equals x1 it equals x1
Write the equation in slope intercept form of the line that has a slope of 2 and contains the point 3 7?
y-y1=m(x-x1) y-7=2(x-3) y-7=2x-6 y=2x-1 y=2x-1
What equation has a graph perpendicular to y equals 2x-3?
5
What is the equation of the line that goes through -1 3 and 5 -9?
(-1,3),(5,-9) (y2-y1)/(x2-x1)= (-9-3)/(5--1)= (-12/6)= the slope is -2 m(x-x1)=y-y1 -2(x-5)=y--9 -2x+10=y+9 y=-2x+1
Determine if the point 3-4 is a solution for the equation y equals 8 -4x?
use the formula y-y1=m(x-x1)
