The highest degree term isn't constant.
It is a variable in the third degree, (i.e. its value depends on a variable, x, which is then cubed) which in this case is then multiplied by a constant, 5.
This was given in the initial question you posted:
Can't you use basic trig to find that the length of the hypotenuse of a 3 - 1 sided right triangle is 12 + 32 = 10, so the hypotenuse is 101/2.
Then you can use the cosine or sine rule? Maybe the cosine is best, with three sides known and one angle wanted?
Realised this about an hour after posting this morning!
As tiny-tim says, draw the triangle remembering that tan is opposite over adjacent. Use Pythagoras and basic trig to calculate the angle!
Are the possible answers very far apart so there is only one logical answer?
I don't know how this would be solved without a calculator or log tables unless this is the case.
Oh, also if you use a calculator, is the correct answer on the list of possible answers? Just to check if you've...
For your second problem, think of the definition of a ring. It is a group, with two operations (R,+,.) ... What do you know about the operation of addition in the ring? i.e. what must (R,+) be?
Think of this problem in two parts, you're already going along the right lines. Take the second part of the problem first. An object (here two cars together) of mass m1 and m2 moves with an initial velocity v. This is what you need to find. For this motion, you know the mass of the object and...
Check your calculations again maybe - I got a different time. Intuitively, 1.9s is far too short. What does your eqn look like when you try to solve for t? How did you find your values?
Think about the motion of the two cars joined together - treat this as a single object which moves in the direction and distance that is stated in the problem. What forces are acting on this object? It is moving with a certain velocity v, then it moves 3.705m, and final velocity is 0 when it's...
For the second part of the race, the eight seconds at constant velocity, just think displacement = velocity X time.
Your v=d/t = 100/10 isn't correct because velocity wasn't constant for the entire 100m.
For the first two seconds, acceleration was constant, and velocity was increasing (at a...
Which direction is friction acting in?
I think you have it right in your picture, but you use it incorrectly when you're resolving forces in your equation.
Consider the 100m broken down into two parts
- the part with constant acceleration from part a). You were asked to get the displacement of this in terms of A in part a). This will help you with that.
- and consider the second part where velocity is constant. what formulae can you use?
For the second question, 26/52 is the probability of a red card, not of a "red heart".
All hearts are red, so not sure if they mean "what is the probability of getting two hearts successively". If it does, your numbers are wrong.
Why are you using a common value for the velocities after the collision? (v1)
That's generally for an inelastic collision, where both objects 'stick together' and move off with a common v.
Here the collision is perfectly elastic.
What do you know about perfectly elastic collisions? What...
Shouldn't the 5-x term be in the numerator and the 3-x term in the denominator?
I'm not 100% certain, but I thought log b - log a = log (b/a), not log (a/b)?
By minimum area, do you mean the sum of the area of the six surfaces? Are there six, is the top included?
Often these questions are asked to minimise material for packaging to hold a certain volume which I'd say this question is about. Not sure though...