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Dillon M.
Engineering Student at SDSU
Tutor Satisfaction Guarantee
Pre-Calculus
TutorMe
Question:

Find the zeros of the following polynomial function. $$F(x) = x^3 +7x^2 +12x$$

Dillon M.
Answer:

First, see if any of the components of the problem share a variable. In this situation they all share an x, allowing it to be factored out. $$F(x) = x(x^2 + 7x+ 12)$$ Now, try to factor out the exponents. $$F(x) = x((x+4)(x+3))$$ Based off of this factoring, one can distinguish that -4, -3, and 0 are all zeros because: 0 = -4 -4 0 = -3 -3 0 = 0 - 0 zeros = -4, -3, 0

Mechanical Engineering
TutorMe
Question:

A hook is drilled into a wall and has two cables attached to it pulling away from it. If one cable pulls away from the wall 300N at an angle of 40deg from the horizontal of another pulls 200N -25deg from the horizontal, what is the magnitude and direction of the resultant vector. If the cement it is drilled into has a critical strength of 500N, does the system fail?

Dillon M.
Answer:

By starting both vectors at (0, 0), one can use Cartesian vector combination by deriving the x and y components of each vector and adding them: $$ F1 = (300cos(40))i + (300sin(40))j$$ $$F2 = (200cos(25))i - (200sin(25))j$$ here you combine the i components and combine the j components to create one vector $$ F(r) = [300cos(40) +200cos(25)]i + [300sin(40) - 200sin(25)]$$ $$ F(r) = 411.1i + 108.3j$$ $$\mid F(r) \mid = \sqrt{411.1^2 +108.3^2}$$ $$\mid F(r) \mid = 425.2N$$ $$\Theta = tan(108.3 / 411.1)$$ $$\Theta = 27deg$$ F(fail) < F(r), therefore the system does not fail

Calculus
TutorMe
Question:

Solve the given integral. f(x) = $$\int_0^{\pi/3}xcos(4x)dx$$

Dillon M.
Answer:

The first step to this problem is identifying that you are going to use the integration by parts method for solving this problem. You know to do this because the x outside of the cosine function makes u-substitution impossible. For integration by parts, one must determine which part of the function one wants to be the "u" and which to be the "dv". The basic rule is the more complicated of the two should be the dv and this stands true in this problem: $$u = x, dv = cos(4x)$$ Once these pieces have been identified one must solve u for du and integrate dv for v: $$u = x$$ $$dv = xcos(4x)$$ $$du = dx$$ $$v=(1/4)sin(4x)$$ With these pieces solved for one can insert them into the integration by parts equation and solve: $$\int_a^b{u} dv = uv - \int_a^b{vdu}$$ $$\int_0^{\pi/3}xcos(4x)dx = [x(1/4)sin(4x)]_0^{\pi/3}-\int_0^{\pi/3}(1/4)sin(4x)dx$$ $$ = [(x/4)sin4x ]_0^{\pi/3}+[(1/16)cos(4x)]_0^{\pi/3}$$ Next, plug the limit values in for x in the form of $$ f(b) - f(a) = f(x)]_a^b$$ f(x) = $$ =(\pi/12)sin(4\pi/3) + (\pi/48)cos(4\pi/3)$$ ...and solve $$f(x) = (\pi/12)(-\sqrt3/2) + (\pi/48)(-1/2) $$ $$ f(x) = (-\pi\sqrt3/24) - (\pi/96)$$

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