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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.
Answer:

#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.
Answer:

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.
Answer:

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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