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Michael L.
Aerospace Software Engineer
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C++ Programming
TutorMe
Question:

A quadrilateral is a shape that has 4 angles that add up to 360. Write a program that is given 3 angles of a quadrilateral and computes the 4th.

Michael L.

#include <iostream> using namespace std; int main() { float ang1, ang2, ang3, ang4; cout << "\n\n Find the fourth angle of a quadrilateral:\n"; cout << "-----------------------------------------\n"; cout<<" Input the 1st angle of the quadrilateral: "; cin>>ang1; cout<<" Input the 2nd angle of the quadrilateral: "; cin>>ang2; cout<<" Input the 3rd angle of the quadrilateral: "; cin>>ang3; ang4=360-(ang1+ang2+ang3); cout << " The 4th angle of the quadrilateral is : " << ang4 << endl; cout << endl; return 0; }

Algebra
TutorMe
Question:

Find the $$x$$ and $$y$$ intercepts for the equation $$y = 3x + 4$$.

Michael L.

To find the y-intercept, we set $$x = 0$$ in our equation and solve for $$y$$. $$y = 3\times{0} + 4$$ $$y = 0 + 4$$ $$y = 4$$ This gives us our y-intercept $$y = 4$$. To find the x-intercept, now we set $$y = 0$$ and solve for x. $$0 = 3x + 4$$ We first subtract $$4$$ from both sides of the equation. $$0 = 3x + 4$$ $$-4\hspace{.7cm} -4$$ $$-4 = 3x$$ Now we can divide both sides by 3 to isolate x. $$\frac{-4}{3} = \frac{3x}{3}$$ $$-\frac{4}{3} = x$$ This gives us our x-intercept $$x = -\frac{4}{3}$$.

Physics
TutorMe
Question:

A car traveling in a straight line crashes into a stationary car in a perfectly inelastic collision. Immediately after the collision, the cars are traveling at $$80 \frac{m}{s}$$. Assuming both cars are the same mass, how fast was the car that crashed traveling immediately before the crash?

Michael L.

To answer this problem, we can use conservation of momentum. $$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$$ We're told that the cars are of the same mass, which means $$m_{1} = m_{2}$$, so we can write all the masses as simply $$m$$. $$mv_{1_{i}} + mv_{2_{i}} = mv_{1_{f}} + mv_{2_{f}}$$ Since the collision is perfectly inelastic, we know that the two cars will travel together at the same speed, which means $$v_{1_{f}} = v_{2_{f}}$$. This lets us simplify the right hand side of our equation as $$mv_{1_{i}} + mv_{2_{i}} = 2mv_{f}$$. Then, our conservation of momentum equation becomes $$mv_{1_{i}} + mv_{2_{i}} = 2mv_{f}$$ Now we can simplify our equation even more. We're told that the car that's hit is initially at rest, so $$v_{2_{i}} = 0$$. We can also divide the mass $$m$$ on both sides, since it appears in every term. Now we're left with $$v_{1_{i}} = 2v_{f}$$ Our final step is to plug in the value that we're given for the final velocity $$v_{f}$$, which is $$80 \frac{m}{s}$$. Plugging that in and solving for $$v_{1_{i}}$$ gives us our answer $$v_{1_{i}} = 160 \frac{m}{s}$$.

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