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4y^2 = 256, Find a value for y

y^2 = 256/4 = 64
therefore, y = sqrt(64) = 8

KP
Answered by Kai P. Maths tutor
3012 Views

A circle with centre C has equation x^2+8x+y^2-12y=12. The points P and Q lie on the circle. The origin is the midpoint of the chord PQ. Show that PQ has length nsqrt(3) , where n is an integer.

First complete the square for both x and y. Move all constants to the right hand side. The square root of this is the radius of the circle. The two constants in the completed square bracket show the x and...

Answered by Maths tutor
7883 Views

Solve the simultaneous equations: 2x + y = 12; x - y = 6

We add the two equations together (left-hand sides and right-hand sides separately). By doing this we get: 2x + y + (x - y) = 12 + 6. By rearranging and simplifying: 3x = 18.If we divide both sides by 3 w...

RD
Answered by Rebeka D. Maths tutor
5494 Views

Integrate(1+x)/((1-x^2)(2x+1)) with respect to x.

To simplify the fraction first notice (1-x2) = (1-x)(1+x) so the common factor of (1+x) in the numerator and denominator can be cancelled. (1+x)/((1-x^2)(2x+1)) = 1/((1-x)(2x+1)), then we need ...

RH
Answered by Ravinder H. Maths tutor
3440 Views

What is 7 to the power of 8? (

Rearrange to (74)2 = ((72)2)272 = 49(72)2 = 492 = (50-1)2 = 502 - 2x50 + 1 = 24017<...

MH
Answered by Michael H. Maths tutor
4382 Views

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