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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
2636 Views

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

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

Answered by Megan G. Maths tutor
2938 Views

Solve the simultaneous equations x^2 + y^2 =13 and x= y - 5.

(y-5)2 + y2 = 13y2-10y +25 + y2=13.2y2-10y+12=0.y2-5y+6=0.(y-3)(y-2)=0. So, y=3 or y=2. Then substituting these values of y back into x=y-...

Answered by Hollie D. Maths tutor
4215 Views

Find the exact value of the integral of (2+7/x), between x=1 and x=e. Give your answer in terms of e.

1.) Find the integral of each term. --> [2x +7ln(x)]. --> Uses standard integrals--> e.g. that the integral of 1/x is ln(x).
2.) substitute values into the integral. --> [2(e)+ 7ln(e)]-...

Answered by Maths tutor
3105 Views

Expand (x+4)(x-4)=33 to give values of x

Use FOIL to expand the brackets. To do this, we start by multiplying the FIRST value in each bracket which in this case is x so we get x^2. Then we multiply the OUTSIDE two values, so here that is x and -...

Answered by Aria B. Maths tutor
2588 Views

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