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Simplify 3log(x^2)+4log(y^3)

Using rules of logarithms: 3log(x^2)+4log(y^3)=log((x^2)^3)+log((y^3)^4) =log(x^6)+log(y^12) =log((x^6)(y^12)

Answered by Alexander C. Maths tutor
3725 Views

Solve the equation (6/x-2)-(6/x+1) =1

1 - Multiply each part of equation by denominators to multiply out the fractions. {(6(x-2))/x-2} - {(6(x-2))/x+1} = 1(x-2). This would simply to 6 - {(6(x-2))/x+1} = x - 2. We then do the same for the oth...

Answered by Aman K. Maths tutor
8127 Views

A is the point with coordinates (2,7) and B is the point with coordinates (8,10). Find the coordinates of the midpoint of AB.

1 - Draw a sketch graph of where the points would be to help visualise the midpoint. 2 - Work out the midpoint for each axis. For example, for x, it would be (0.5*(2+8)) = 5 and for y, it would be (0.5*(7...

Answered by Aman K. Maths tutor
4748 Views

(Q20 Non-Calculator paper, Higher Tier) Solve algebraically the simultaneous equations: x^2 + y^2 = 25 and y – 3x = 13

We have a quadratic equation (unknowns are raised to the power of 2) and a liner equation (power 1). We will use the linear equation to express one of the variables, e.g. y, in terms of t...

Answered by Darena S. Maths tutor
4210 Views

A 20-foot ladder is leaning against a vertical wall. The bottom of the ladder is pulled away horizontally from the wall at 3 feet per second. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 10 feet away?

This question can be divided into two different equations. In order to calculate how fast the ladder is sliding downwards, you need to know at what original vertical height the ladder is leaning on the wa...

Answered by Maximilian L. Maths tutor
6077 Views

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