Top answers

differentiate x^2 + y^3 + xy respect to x

2x + 3y 2 * dy/dx + x * dy/dx +y
XJ
Answered by Xianming J. • Maths tutor
3821 Views

dx/dt=-5x/2 t>=0 when x=60 t=0

dx/dt=-5x/2 Int(x, dx)=Int(-5/2, dt) ln(x)=-5t/2+c x=60 when t=0 ln(60)=c ln(x)=ln(60)-5t/2 x=e ln(60)-5t/2 x=60/e 5t/2
FD
Answered by Felix D. • Maths tutor
5174 Views

Solve the simultaneous equations x^2 + y^2 = 9 and y = 3x + 3

y 2 = (3x + 3) 2 = 9x 2 + 18x + 9 x 2 + 9x 2 + 18x + 9 = 9 10x 2 + 18x = 0 (root a: x = 0) 5x + 9 = 0 (root b: x = -1.8) For x = 0, y = 3(0) + 3 (root a: y = 3) For x = -1.8, y = 3(-1.8) + 3 (root b: y = -2....
AR
Answered by Alistair R. • Maths tutor
4547 Views

How would you increase 400 by 7%

First, we'd want to find out what 1% of 400 is. If 400 is the total number, that would make 400 the whole 100%. In order to find out 1% we'd want to divide the number by 100 which is 4. Now that we know 1% i...
TT
Answered by Tharani T. • Maths tutor
14665 Views

Given that sin(x)^2 + cos(x)^2 = 1, show that sec(x)^2 - tan(x)^2 = 1 (2 marks). Hence solve for x: tan(x)^2 + cos(x) = 1, x ≠ (2n + 1)π and -2π < x =< 2π(3 marks)

sin(x) 2 + cos(x) 2 = 1 Dividing by cos(x) 2 gives: tan(x) 2 + 1 = sec(x) 2 Which rearranges as: sec(x) 2 - tan(x) 2 = 1 as required. tan(x) 2 + cos(x) 2 = 1 sec(x) 2 - 1 + cos(x) 2 = 1 sec(x) 2 + cos(x) 2 =...
AR
Answered by Alistair R. • Maths tutor
4371 Views