They are clearly parallel and don't coincide or meet anywhere.Without actually solving the simultaneous equations, state whether the following pairs of lines intersect are parrallel or coincide
a) x+y-7=0 and x+y-8=0
They are clearly parallel and don't coincide or meet anywhere.Without actually solving the simultaneous equations, state whether the following pairs of lines intersect are parrallel or coincide
a) x+y-7=0 and x+y-8=0
Pretty muchWhat about 6x-5y-24=0 and 9x-4y-22=0. Do you so these things by inspection?
Write b = 2a, c = 3a, then substitute into the given equation and solve for a.If a:b:c is 1:2:3 and a^2+b^2+c^2=504 find a, where a is a positive number
Differentiate and find where the derivative is 0 for stationary points. Determine their nature using for instance the second derivative test. Sub. in the x-values of the stationary points to the f(x) expression to get the coordinates.Consider the curve given by y=x^3-6x+4
FInd the coordinates of the stationary points and determine their nature
Find the coordinates of any points of inflection
Sketch the curve for the domain -3=<x=<3
What is the maximum value of x^3-6x+4 in the domain -3=<x=<3
IS this the highest part of the graph?Differentiate and find where the derivative is 0 for stationary points. Determine their nature using for instance the second derivative test. Sub. in the x-values of the stationary points to the f(x) expression to get the coordinates.
Find where f"(x) = 0 for the inflection point.
Last part can be done by inspection after doing the previous parts.
Yeah the max. value is the y-value of the highest point on the graph.IS this the highest part of the graph?
The answer is just the perpendicular bisector of A and B. We can find the equation of this line by finding the midpoint of A and B, and then finding the slope of the line (which will be the negative reciprocal of the slope of AB, since it's perpendicular to AB). Once we've worked out the slope and the midpoint, we can use point-gradient formula to get the equation.
i) (k-1)^2 - 4(2)(2)