3 years ago

Someone please help me??

1 answer:

7 0

The standard equation of a circle with centre (xc,yc) and radius R is given by

$(x-x_c)^2+(y-y_c)^2=R^2$

Substituting
centre (xc,yc) = (4,-3)
R=2.5

The equation is therefore

$(x-x_c)^2+(y-y_c)^2=R^2$
$(x-4)^2+(y-(-3))^2=2.5^2$
$(x-4)^2+(y+3)^2=2.5^2$ or
(x-4)^2+(y+3)^2=6.25

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Solve the initial value problem. dy/dt = 1 + 6/t , t > 0, y = 8 when t = 1

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$\displaystyle\frac{dy}{dt} = 1 + \frac{6}{t}\ \Rightarrow\ dy = \left( 1 + \frac{6}{t}\right) dt\ \Rightarrow\int 1\, dy = \int \left( 1 + \frac{6}{t}\right) dt \ \Rightarrow \\ \\
y = t + 6\ln|t| + C. \text{ But $t\ \textgreater \ 0$ so }y = t + 6\ln t + C. \\ \\ 
y(1) = 8\ \Rightarrow\ 8 = 1 + 6 \ln 1 + C \ \Rightarrow\ C = 7 \text{ so } \\ \\
y(t) = t + 6\ln t + 7$