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This is a linear ODE with $P = 3\sec^2 x$ and $Q = \sec^2 x$.
Integrating factor: $e^{\int 3\sec^2 x\,dx} = e^{3\tan x}$
$\frac{d}{dx}(ye^{3\tan x}) = \sec^2 x \cdot e^{3\tan x}$
Integrate: $ye^{3\tan x} = \frac{1}{3}e^{3\tan x} + C$
At $x=\pi/4$, $\tan(\pi/4)=1$, $y=4/3$: $\frac{4}{3}e^3 = \frac{1}{3}e^3 + C \Rightarrow C = e^3$
So $y = \frac{1}{3} + e^3 \cdot e^{-3\tan x}$
At $x=-\pi/4$, $\tan(-\pi/4)=-1$: $y = \frac{1}{3} + e^3 \cdot e^{3} = \frac{1}{3} + e^3$... wait: $e^{-3(-1)}=e^3$, so $y = \frac{1}{3}+e^3 \cdot e^3 = \frac{1}{3}+e^6$.
Correct: $y(-\pi/4) = \mathbf{\frac{1}{3}+e^6}$. So correct option index is 0.
A ball is dropped from rest at $t = 0$. After 6 s, another ball is thrown downward from the same point with speed $v$. Both balls meet at $t = 18\text{ s}$. Find $v$ ($g = 10\text{ m/s}^2$).
Ball 1 falls for 18 s: $h_1 = \dfrac{1}{2}(10)(18)^2 = 1620\text{ m}$
Ball 2 falls for $18 - 6 = 12\text{ s}$: $h_2 = 12v + \dfrac{1}{2}(10)(12)^2 = 12v + 720$
Setting $h_1 = h_2$: $12v + 720 = 1620 \Rightarrow 12v = 900 \Rightarrow v = \mathbf{75\text{ m/s}}$
Extension $\Delta L = \frac{FL}{AY} = \frac{FL}{\pi (d/2)^2 Y} \propto \frac{L}{d^2}$ [cite: 482, 484].
- New extension $\Delta L' \propto \frac{2L}{(2d)^2} = \frac{2L}{4d^2} = \frac{1}{2} \Delta L$
- $\Delta L' = \frac{1}{2} \times 0.04 = 0.02\text{ m}$[cite: 484].
Given the diagram of three circles with centers $P$, $Q$, and $R$ forming an equilateral triangle $PQR$, where each pair of circles has exactly one point in common:
Compare:
Column A: The perimeter of triangle $PQR$
Column B: The circumference of the circle with center $Q$
Since each pair of circles has exactly one point in common, the circles are externally tangent to each other. If all circles have the same radius $r$, then the distance between any two centers equals $2r$.
Since triangle $PQR$ is equilateral with side length $2r$:
- Perimeter of $\triangle PQR = 3(2r) = 6r$
- Circumference of circle with center $Q = 2\pi r$
Comparing: $6r$ vs $2\pi r$
Dividing by $2r$: $3$ vs $\pi$
Since $\pi \approx 3.14159 > 3$, we have $2\pi r > 6r$.
Therefore, Column B (the circumference) is greater than Column A (the perimeter).
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