Top answers

Maths
All levels

Differentiate x^3⋅cos(5⋅x) with respect to x.

In order to solve this problem we will have to use the product rule as follows: d/dx[x^3⋅cos(5⋅x)]=[d/dx(x^3)]⋅cos(5x)+(x^3)⋅[d/dx[cos(5x)]]=(3⋅x^2)⋅cos(5⋅x)+(x^3)⋅−5⋅sin(5⋅x)=3⋅x^2⋅cos(5⋅x)−5⋅x...

TL
Answered by Tianyu L. Maths tutor
5672 Views

Work out 2 7/15 -1 2/3

First turn into improper fractions37/15 - 5/3Get same denominator on bottom37/15-25/15Take away numerator (37-25)/15= 12/15 Simplify 4/5

HG
Answered by Henry G. Maths tutor
4792 Views

How would you factorise x^2 + 4x + 4

This is in the form ax2 + bx + cWhere, a= 1b=4c=4To factor we have to find two numbers that:Add up to b, in this case 4Multiply together to give c

TB
Answered by Toby B. Maths tutor
2838 Views

Points P and Q are situated at coordinates (5,2) and (-7,8) respectively. Find a) The coordinates of the midpoint M of the line PQ [2 marks] b) The equation of the normal of the line PQ passing through the midpoint M [3 marks]

a) For finding the midpoint M, the point M must be equidistant from P and Q in both the x and y axes. Hence, we consider the x and y axis separately. The midpoint of the x coordinates is essentially a mea...

JU
Answered by Justin U. Maths tutor
3809 Views

The gradient of the curve at A is equal to the gradient of the curve at B. Given that point A has x coordinate 3, find the x coordinate of point B.

We are told f(x) = (2x-5)^2 (x+3).In part b) we are asked to show that f'(x) = 12x^2 -16x -35, so for part (c) we shall assume this definition for f'(x). We are told that the x coordinate for A is 3. Call...

TF
Answered by Thomas F. Maths tutor
5867 Views

We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences