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The area of a parallelogram is given by the equation 2(x)^2+7x-3=0, where x is the length of the base. Find: (a) The equation of the parallelogram in the form a(x+m)^2+n=0. (b) The value of x.

(a)STEP 1: Take out the coefficient of x^2 from the x^2 and x terms.
2(x)2+7x-3=0 2(x2+(7/2)x)-3=0
STEP 2: Complete the Square by finding (b/2)2.
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Answered by Marco A. Maths tutor
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Solve, using the quadratic formula, the equation x^2 +2x=35

suppose you have a general quadratic equation in the form ax2 + bx + c = 0, then the solutions to this quadratic equation can be found by the equation x = (-b +/- (b2-4ac)1/2

Answered by Antonia S. Maths tutor
2588 Views

Given that x = 4sin(2y + 6), Find dy/dx in terms of x

x = 4sin(2y + 6)dx/dy = 4(2)cos(2y + 6)dx/dy = 8cos(2y + 6) ==> dy/dx = 1/8cos(2y + 6)sin2(2y + 6) + cos2(2y + 6) = 1cos(2y + 6) = √(1 - sin2(2y + 6))cos(2y + 6) = √(1 ...

Answered by Shubham P. Maths tutor
7835 Views

Find the area under the curve y=xexp(-x)

first recognise ‘area under’ = integrationfunction is a product so by partsset x=u; exp(-x)=dv/dx integral(udv/dx)=uv-integral(vdu/dx)
du/dx =1; v=-exp(-x)ans=(x.-exp(-x))-integral(1.exp(-x)dx)ans=-x...

Answered by Maths tutor
2523 Views

Prove that 2Sec(x)Cot(x) is identical to 2Cosec(x)

2 Secx = 2/CosxCotx = 1/Tan x = Cosx/SinxTherefore: 2SecCotx = 2/Cosx * Cosx/Sinx = 2/Sinx = 2Cosecx

Answered by Ben S. Maths tutor
3532 Views

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