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a curve has an equation: y = x^2 - 2x - 24x^0.5 x>0 find dy/dx and d^2y/dx^2

dy/dx = 2x -2 - 12x^-0.5d^2/dx^2 = 2 + 6 x^-3/2

Answered by Maths tutor
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Solve (x-1)^2-2(x-1)-3=0

Solving this equation would mean finding the values of x that satisfy it equating to zero. Therefore, to start, you must simplify the equation by expanding all the values to get (x^2-2x+1)-(2x-2)-3=0 [exp...

EL
Answered by Emily L. Maths tutor
3631 Views

Solve the equation x^6 + 26x^3 − 27 = 0

While this equation may look complicated, it's actually much easier than it looks - this equation is called a hidden quadratic. This is because it can be rewritten and solved the same way you would solve ...

SP
Answered by Saachi P. Maths tutor
4926 Views

Find the equation of the normal to the curve 2x^3+3xy+2/y=0 at the point (1,-1)

Step 1 Use Implicit differentiation with respect to x and y - 6x^2+3y + 3x(dy/dx) - 2/y^2(dy/dx) = 0Step 2 Write the equation as dy/dx =... - dy/dx = (6x^2 + 3y)/(2/y^2 - 3x)Step 3 Input (1,-1) into th...

Answered by Maths tutor
2944 Views

Solve the following equation: 13y - 5 = 9y + 27

13y - 5 = 9y + 2713y - 9y = 27 + 54y = 32y = 32/4y = 8

MG
Answered by Megan G. Maths tutor
3258 Views

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