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A tunnel has height, h, (in metres) given by h=14-x^2 where x is the horizontal distance from the centre of the tunnel. Find the cross sectional area of the tunnel. Also find the maximum height of a truck passing through the tunnel that is 4m wide.

Firstly, solve 0=14-x^2 to find the horisontal distance to the edges of the tunnel. x1=sqrt(14), x2= -sqrt(14). Integrate h=14-x^2 between x1 and x2 28*sqrt(14) -(2(sqrt(14)^3))/3. This is the required area ...
JG
Answered by James G. • Maths tutor
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Express 3(x^2) - 12x + 5 in the form a(x - b)^2 - c.

Starting with a(x - b)^2 - c, if we expand the bracket we get: a(x^2 - 2xb + b^2) -c Since we need to end up with the coefficient on x^2 being 3 and in the expression above x^2 is only multiplied by a, this ...
LH
Answered by Lucy H. • Maths tutor
11005 Views

Make x the subject of the equation. y = 4( 2 + x )/ (6x -1)

First lets times both sides by (6x -1) y(6x - 1) = 4(2 + x) now lets expand the brackets 6xy - y = 8 + 4x now lets move all the ones with x as a factor onto one side of the equation and the rest on the other...
OO
Answered by Oliver O. • Maths tutor
5957 Views

Calculate the area of a circle of diameter 8cm

Solution: We can first write down the formula for the area of a circle which is pi*(radius)^2. The radius of the circle is half the diameter, so radius= 1/2(8)=4cm. Therefore the area of the circle is: pi* (...
HH
Answered by Headon H. • Maths tutor
20949 Views

Write the complex number Z=1/2+sqrt(3)/2j both as a function involving cos & sin, and as a function involving an exponential.

|Z| = sqrt(1/2^2 + (sqrt(3)/2)^2) = 1 arg(Z) = arctan((sqrt(3)/2)/(1/2)) = pi/3 Z = cos(pi/3) + jsin(pi/3) Z = e^j(pi/3) Apologies for the use of sqrt(), I have no way yet of using the symbol on my laptop
SR
Answered by Sol R. • Maths tutor
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