Tutor profile: Azin E.
what is the lim [e^x -1] / x as x approaches 0
We could write the above statement as: lim [e^x - e^0] / x-0 as x approaches 0. The definition of the derivative at x = a is given by f '(a) = lim [f(x) - f(a)] / (x - a) as x approaches a. in this case, we have the above mathematical statement and we have f(x) =e^x The given limit is the derivative of e^x at x = 0, since we know that f'(x) for an exponential function is also e^x, the derivative of e^x at x=0 is e^0 = 1. Any questions?
Two forces are acting on a 5.00 kg mass. One of the forces is 10.0 N south and the other is 15.0 N east. The magnitude of the acceleration of the mass is:
So we see that one of the forces is to the south and the other is to the east, let's assume the north is upward, the south is downward, the east is to the right and the west is to the left, so we have one downward force and one force to the right, they are perpendicular. If we define our x-y plane to be such that our x-axis is east-west and our y-axis is north-south we could write the second law of Newton as: sigma_y(f) = ma_y ==> 10 N= 5 * a_y ==> a_y= 2 (1) and sigma_x (f)=ma_x ==> 15 N =5 * a_x ==> a_x= 3 (2) Now we know that the acceleration is a vector and in this case, the overall acceleration is in the south-east direction, since we have components of it in south and east directions. and according to the Pythagorean theorem, the magnitude of the overall acceleration then should be: the square root of ((2)^2 + (3)^2) = square root of 13, which is 3.6 if you try it in your calculator. Any questions?
My father’s age divided by 5 is equal to my brother’s age divided by 3. My brother is 3 years older than me. My father’s age is 3 less than 2 times my age. How old is my father?
Let's show my father's age by the letter x, my brother the letter y and my age by letter z. From the question, we could see that we need to find my father's age, x. In the question, we are told that my father's age divided by 5 (x/5) equals to my brother's age divided by 3, (y/3) so: x/5 = y/3. we also know, my brother is 3 years older than me, so: z+3=y. and my father is 3 years younger than 2 times my age so if I double my age and subtract 3 from it, I will have my father's age: 2z-3=y so let's summarize what we have: x/5=y/3 (1) z+3=y (2) 2z-3=x (3) from equation 3, we can see that 2z-3=x, so we could put (2z-3) instead of x in equation 1: (2z-3)/5=y/3 and let's multiply both sides of the equation by 15, so we could get rid of the fractions, note that we choose 15 because it is the lowest number that is divisible by both 3 and 5. so now we would have 6z-9=5y (1) z+3=y (2) Now let's put z+3 instead of y in equation 1 since we have this in equation 2: 6z-9=5(z+3) ==> 6z-9 = 5z+15 let's subtract 5z from both sides of the equation: z-9=15 ==> z= 15+9 (bringing 9 to the other side of the equation, as if we are adding +9 to both sides) ==> z=24 so I am 24 years old. Now I know my brother is 3 years older so he is 27 years old. and according to equation 3, 2z-3=x so 24*2 -3 = 45 is my father's age. So the key thing in solving such complicated algebraic equations is to try to write all the terms in terms of one other thing, solve for one of the unknowns, like we first wrote y in terms of z and then we did write x in terms of z, because in equation 2 and 3, we had everything in terms of z. Then our equation 1, became an equation with one unknown z, and then after finding z, we could use other equations, in this case, 2 and 3 to find our x and y. I hope it's clear for you. Any questions? :)
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