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2015年11月13日 · This is the equation of a circle of radius 4 with centre (0, 0). If you attempt to define y in terms of x you get multiple values for x in (-4, 4). Starting with x^2+y^2=16, subtract x^2 from both sides to get: y^2 = 16-x^2 Then taking square roots of both sides: y = +-sqrt(16-x^2) = +-sqrt(4^2-x^2) Now sqrt(4^2-x^2) only takes Real values when 4^2-x^2 >= 0, that is when …

c = 16 Substitute these values into the formula above to find that (0, 16) is the vertex point. The expression 16 - x^2 can be factored as a difference of squares. (4 + x)*(4 - x)=0 So, the x-intercepts of the graph are (-4, 0) and (4, 0). Pick some other values of x, and find the corresponding values of y to get a few more points to plot.

2018年3月23日 · Perform the substitution. #x=4sintheta#, #=>#, #dx=4costhetad theta# #sqrt(16-x^2)=sqrt(16-16sin^2theta)=4costheta#

2018年2月22日 · If x^2+16=0 x=+-sqrt(-16) If ax^2+bx+c=0, then x=1/(2a)(-b+-sqrt(b^2-4ac)) For x^2+16=0 a=1, b=0, c=16 Substituting x=1/(2xx1)(-0+-sqrt(0^2-4xx1xx16)) =1/2xx+-sqrt ...

\[\int \frac{\sqrt{{x}^{2}-16}}{x} \, dx\] +. > < ...

\[{x}^{2}-16\] +. > < ...

2018年6月1日 · Substitute #x=4sin theta# to turn the denominator into a simple trig function via the identity #sin^2 theta +cos^2 theta=1#. Note that #(dx)/(d theta)=4 cos theta# and that the limits of integration #x=0,2sqrt(3)# are equivalent to #theta=0,pi/3# .

2018年4月7日 · Here, #I=intx^3sqrt(16-x^2)dx=intx^2sqrt(16-x^2)*xdx# Let, #x=4sinu=>x^2=16sin^2u=>2xdx=32sinucosudu# i.e. #xdx=16sinucosudu#

2018年5月5日 · Here, #I=intx^3sqrt(16-x^2)dx# Let, #x=4sinu=>dx=4cosudu# #and sinu=x/4=>cosu=sqrt(1-sin^2u)=sqrt(1-x^2/16# ...