1) How do you express a vector as a linear combination of i and j unit vectors?
For example, v = <2,-2> and w = <0,3>.
2) How do you solve tricky radical equations like this? When trying to repeatedly square both sides to get rid of the root, the equation doesn’t check.
I.E. sqrt(5x-3) – 2sqrt(x)=6
3) When determining profit from a cost equation and a revenue equation, do you set both equations to be equal to each other in order to find the break even point?
C = 70 – 11p
R = 14p – 2p^2
4) How do you solve questions of force and magnitude using vectors? I learned about solving questions of direction and heading using vectors, but never force.
For example, a truck weighing 6578 pounds is on a driveway inclined 17.5 degrees to the ground. Neglecting friction, what’s the magnitude of the force parallel to the driveway that will keep the truck from rolling down the incline?
Thanks in advance
Vector v = 2i – 2j
Vector w = 0i + 3j = 3j
sqrt(5x-3) – 2sqrt(x)=6
Yes, square both sides….
(5x-3) -4sqrt(5x^2 – 3x) + 4x = 36
9x -39 = 4sqrt(5x^2-3x)
square both sides again…
(9x -39)(9x-39) = 16(5x^2 – 3x)
81x^2 – 702x + 1521 = 80x^2 – 48x
x^2 -654x + 1521 = 0
factor…. and solve for “x”…. If this is not factorable, then you have to use the quadratic equation to find the roots…
Yes… to find the break even point… you set the revenue equation “equal to” the Cost equation…
A FORCE is a vector.
a vector is just the mathematical representation of something with magnitude and direction.
the weight is a vector 6578 lb straight down.
you can pull allong the incline at 17.5 degrees above horizontal.
how much of the 6578 is parralel to the drive….
break up 6578 lb into two components, one parallel to the slope, and the other perpendicular to the slope.
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