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Before Maria changed jobs, her salary was 24 percent more than Julio's salary. After Maria changed jobs, her new salary was 24 percent less than her old salary.
Compare:
Column A: Julio's salary
Column B: Maria's new salary
Let Julio's salary be $J$ and Maria's old salary be $M_{old}$.
Given: $M_{old} = 1.24J$
Maria's new salary: $M_{new} = M_{old} - 0.24M_{old} = 0.76M_{old} = 0.76(1.24J) = 0.9424J$
Comparing:
- Column A: $J = 1.0J$
- Column B: $0.9424J$
Since $1.0 > 0.9424$, Column A (Julio's salary) is greater than Column B (Maria's new salary).
Therefore, Column A is greater.
Given $x=\sin^{-1}(\sin 10)$ and $y=\cos^{-1}(\cos 10)$, find $y-x$.
Note $10$ radians. Since $3\pi\approx9.42$ and $4\pi\approx12.57$: $3\pi<10<4\pi$.
For $x$: $\sin^{-1}(\sin10)$. Principal range $[-\pi/2,\pi/2]$. $10\approx10-3\pi\approx0.58$, but $10-(3\pi)\approx0.58<\pi/2$? $3\pi\approx9.42$, $10-3\pi\approx0.58$. Since $\sin(10)=\sin(\pi-(10-3\pi))=\sin(4\pi-10)$... more carefully: $10 = 3\pi + (10-3\pi)$, $\sin(10)=\sin(3\pi+(10-3\pi))=-\sin(10-3\pi)$. So $x=\sin^{-1}(-\sin(10-3\pi))=-(10-3\pi)=3\pi-10$.
For $y$: $\cos^{-1}(\cos10)$, range $[0,\pi]$. $10-3\pi\approx0.58\in[0,\pi]$, so $y=10-3\pi$... wait: $\cos(10)=\cos(4\pi-10)$ and $4\pi-10\in[0,\pi]$. So $y=4\pi-10$.
$y-x=(4\pi-10)-(3\pi-10)=\mathbf{\pi}$
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