Top answers

Split (3x-4)/(x+2)(x-3) into partial fractions

(3x-4)/(x+2)(x-3) = A/(x+2) + B/(x-3)=> 3x-4 = A(x-3) + B(x+2)Let x = -2-10 = -5AA = 2Let x = 35 = 5BB = 1∴ (3x-4)/(x+2)(x-3) = 2/(x+2) + 1/(x-3)
Answered by • Maths tutor
4050 Views

A circle with centre C has equation x^2+8x+y^2-12y=12. The points P and Q lie on the circle. The origin is the midpoint of the chord PQ. Show that PQ has length nsqrt(3) , where n is an integer.

First complete the square for both x and y. Move all constants to the right hand side. The square root of this is the radius of the circle. The two constants in the completed square bracket show the x and y ...
Answered by • Maths tutor
9124 Views

Integrate(1+x)/((1-x^2)(2x+1)) with respect to x.

To simplify the fraction first notice (1-x 2 ) = (1-x)(1+x) so the common factor of (1+x) in the numerator and denominator can be cancelled. (1+x)/((1-x^2)(2x+1)) = 1/((1-x)(2x+1)), then we need to split thi...
RH
Answered by Ravinder H. • Maths tutor
4318 Views

What is 7 to the power of 8? (

Rearrange to (7 4 ) 2 = ((7 2 ) 2 ) 2 7 2 = 49(7 2 ) 2 = 49 2 = (50-1) 2 = 50 2 - 2x50 + 1 = 24017 8 = (7 4 ) 2 =((7 2 ) 2 ) 2 = 2401 2 = (2000 + 400 + 1) 2 = 5,764,801
MH
Answered by Michael H. • Maths tutor
5812 Views

The numbers a, b, c and d satisfy the following equations: a + b + 3c + 4d = k; 5a = 3b = 2c = d. What is the smallest value for k for which a, b, c and d are all positive integers

5a = 3b = 2c = d. d must be a multiple of 5, 3 and 2, therefore the smallest possible value for d is 30. This sets a = 6, b = 10 and c = 152) a + b + 3c + 4d = 6 + 10 + 3x15 + 4x30 = 181 k = 181
MH
Answered by Michael H. • Maths tutor
4930 Views