Let the slope of the line 45 x+5 y+3=0 be 27 | vedclass

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(B) The equation of the line is $45 x+5 y+3=0$,which can be written as $5 y = -45 x - 3$,or $y = -9 x - \frac{3}{5}$. The slope is $-9$.Given that the

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(B) The equation of the line is $45 x+5 y+3=0$,which can be written as $5 y = -45 x - 3$,or $y = -9 x - \frac{3}{5}$. The slope is $-9$.Given that the slope is $27 r_1+\frac{9 r_2}{2}$,we have $27 r_1+\frac{9 r_2}{2} = -9$.Let $f(x) = \int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} dt$. We want to find $\lim_{x ightarrow 3} f(x)$.Since the limit is of the form $\frac{0}{0}$,we apply $L$'$H$ôpital's Rule:$\lim_{x ightarrow 3} \frac{\frac{d}{dx} \int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} dt}{\frac{d}{dx} (	ext{denominator})}$.Using Leibniz's rule for the numerator,$\frac{d}{dx} \int_3^x g(t, x) dt = g(x, x) + \int_3^x \frac{\partial}{\partial x} g(t, x) dt$. At $x=3$,the integral part vanishes.Thus,the limit becomes $\lim_{x ightarrow 3} \frac{8 x^2}{\frac{3 r_2}{2} - 2 r_2 x - 3 r_1 x^2 - 3}$.Substituting $x=3$: $\frac{8(9)}{\frac{3 r_2}{2} - 6 r_2 - 27 r_1 - 3} = \frac{72}{-\frac{9 r_2}{2} - 27 r_1 - 3}$.Since $27 r_1 + \frac{9 r_2}{2} = -9$,the denominator is $-(-9) - 3 = 9 - 3 = 6$.Therefore,the limit is $\frac{72}{6} = 12$.

Let the slope of the line $45 x+5 y+3=0$ be $27 r_1+\frac{9 r_2}{2}$ for some $r_1, r_2 \in R$. Then $\lim_{x \rightarrow 3} \left( \int_3^x \frac{8 t^2}{\frac{3 r_2 x}{2}-r_2 x^2-r_1 x^3-3 x} dt \right)$ is equal to ...................

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