Angle at centre = 2 x angle at circumference.
since angle BDC is 92 degrees, angle BAC will be 92/2 degrees
Area = b*h and in some cases * 2 so you need to do this, 64*52 + 40*28= 4448 yd2
If A=w(50-w)
A=50w-w^2
dA/dw=50-2w
d2A/dw2=-2
Since the acceleration is a constant negative, that means that when velocity, dA/dw=0, it is at an absolute maximum for A(w)...
dA/d2=0 only when 50=2w, w=25
So as the case with any rectangle, the perfect square will enclose the greatest area possible with respect to a given amount of material to enclose that area...
So the greatest area occurs when W=L=25 in this case:
A(25)=50w-w^2
Area maximum is thus:
Amax=50(25)-(25)^2=625 u^2
A crest a scatteroy displaying the data in the table. Be sure to include a linear trend line
12/28 is equal to 3/7 (divide both the numerator and the denominator by the highest common factor).
