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The average number of pages in 1940 was 30. Twice this amount is $2 \times 30 = 60$ pages.
Check each year:
- 1940: 30 pages (not $\geq 60$) β
- 1950: 55 pages (not $\geq 60$) β
- 1960: 82 pages ($\geq 60$) β
- 1970: 70 pages ($\geq 60$) β
- 1980: 74 pages ($\geq 60$) β
Three years (1960, 1970, and 1980) had at least twice the 1940 average.
$E_1 = \frac{GMm}{R_E} - \frac{GMm}{R_E+h} = GMm\frac{h}{R_E(R_E+h)}$
$E_2 = \frac{GMm}{2(R_E+h)}$ (orbital KE)
Setting $E_1 = E_2$:
$\frac{h}{R_E(R_E+h)} = \frac{1}{2(R_E+h)}$
$\frac{h}{R_E} = \frac{1}{2}$
$h = \frac{R_E}{2} = \frac{6.4 \times 10^3}{2} = 3.2 \times 10^3\ \text{km}$
For each $x\in\mathbb{R}$, let $[x]$ be the greatest integer $\leq x$. Find: $\displaystyle\lim_{x\to0^-}\dfrac{x([x]+|x|)\sin[x]}{|x|}$
For $x\to0^-$: $x<0$, $|x|=-x$, $[x]=-1$.
$\dfrac{x((-1)+(-x))\sin(-1)}{-x} = \dfrac{x(-1-x)(-\sin1)}{-x}$
$= \dfrac{x(1+x)\sin1}{x} = (1+x)\sin1$
As $x\to0^-$: limit $= 1\cdot\sin1$... wait: $= \dfrac{x(-1-x)(-\sin1)}{-x} = \dfrac{(1+x)\sin1 \cdot x}{x}=(1+x)\sin1\to\sin1$? Re-doing: numerator $= x(-1-x)\sin(-1)=x(-1-x)(-\sin1)=x(1+x)\sin1$; denominator $=-x$. So $=\dfrac{x(1+x)\sin1}{-x}=-(1+x)\sin1\to-\sin1$. Answer: $\mathbf{-\sin 1}$.
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