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To differentiate, we bring the power down and decrease the power by 1. So x3 becomes 3x2, -3x2 becomes -6x, and -9x (which can be written as -9x1 ) becomes -9. ...
Area of a circle = pi × r2
Area of semi-circle = 1/2(Area of circle)
= 1/2(pi × r2)
Diameter = 2r
r = 8÷2 = 4
Area of semi-circle = 1/2(pi × 4APAnswered by Annabelle P. • Maths tutor8626 Views
This type of question appears over-complicated with limited options, but one must not fear! At first, the numerator seems to be of a higher degree than the denominator (x^4 compared to x^2), but from the ...
Using the product rule, f=uv, df = (vu'-uv')/v^2. we first set u = 3x^2 and v = sin(2x). u' = 6x, v'=2cos(2x) Therefore, vu' = 6xsin(2x). uv' = 6x^2cos(2x), v^2 = 4cos^2(2x) Therefore the differe...
To answer this question, we need to make y the subject of the second equation. We can do this through simple rearrangement:
y-3x=13 so y=13+3x
Now that we have y on its own, this means we ca...
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