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

(y-5) 2 + y 2 = 13y 2 -10y +25 + y 2 =13.2y 2 -10y+12=0.y 2 -5y+6=0.(y-3)(y-2)=0. So, y=3 or y=2. Then substituting these values of y back into x=y-5 gives, x=-2 when y=3 and x=-3 when y=2.
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Answered by Hollie D. • Maths tutor
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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)]- [2(1)+7ln...
Answered by • Maths tutor
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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 -4 s...
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Answered by Aria B. • Maths tutor
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How do surds relate to powers and roots?

To understand surds we have to understand powers and there uses. So to start we should look at powers such as what is 2 2 , 3 2 , and 5 2 . This shows us going up but then we look at how we square root thing...
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Answered by Alistair G. • Maths tutor
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The velocity of a car at time, ts^-1, during the first 20 s of its journey, is given by v = kt + 0.03t^2, where k is a constant. When t = 20 the acceleration of the car is 1.3ms^-2, what is the value of k?

Our task is to find out the value of k, which we can determine from the equations for velocity or acceleration if we know 2 of the 3 variables in either equation. We are given the value of acceleration at t(...
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Answered by Lascelle M. • Maths tutor
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