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A particle of mass 5kg is held at rests on a slope inclined at 30 degrees to the horizontal. The coefficient of friction for the slope is 0.7, determine whether the particle will move when released.

Frictional force acting:FMax= uR = 0.7 x 5 x 9.81 x cos(30) FMax = 29.7NWeight acting down the slope:F = 5 x 9.81 x sin(30)F = 24.5NFric...

MI
Answered by Myles I. Maths tutor
5106 Views

Find values of x in the interval 0<x<360 degrees. For which 5sin^2(x) + 5 sin(x) +4 cos^2(x)=0

This question is split up into two parts.
Firstly recall the trigonometric identities you know, the trick here is to eliminate one of the squared terms. Using 4sin^2(x) +4cos^2(x) = 4, the cos term i...

JG
Answered by James G. Maths tutor
8954 Views

Here are four fractions, 4/5 3/8 12/30 14/20. Write them in order of size, starting with the smallest fraction.

First cancel down the fractions to make the numbers as small as possible, then make all fractions have the same denominator. 12/30 becomes 2/5. 14/20 becomes 7/10. So the new fractions are 4/5 3/8 2/5 and...

KS
Answered by Kim S. Maths tutor
4835 Views

Differentiate the following... f(x)= 5x^4 +16x^2+ 4x + 5

The rule for differentiating a simple equation such as this is, to times the coefficient by the power and the take one away from the power. Therefore, lets look at 5x^4 you would times the 5 by 4 to get 2...

CM
Answered by Celia M. Maths tutor
3481 Views

Let f be a function of a real variable into the real domain : f(x) = x^2 - 2*x + 1. Find the roots and the extremum of the function f.

Let f be a function of a real variable into the real domain : f(x) = x^2 - 2x + 1. Find the roots and the extremum of the function f.i) Root finding The function f is a 2nd degree polynomial. The root...

MM
Answered by Matteo M. Maths tutor
3072 Views

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