Need answers for questions 1-4. Will give brainless!!

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

4 0

There appear to be 2 treasures marked with X on the map. One is at (15°S, 120°W); the other is at (45°S, 120°E). The first of these is both northernmost and westernmost. The second of these is on its own island. Since it is the only treasure that is not the northernmost, it is also the closest to the northernmost.

(15°S, 120°W)
(15°S, 120°W)
(45°S, 120°E)
(45°S, 120°E)

_____

<em>Comment on the map</em>

There appear to be additional features marked with a double "circle". We cannot tell if these are mountains, or are supposed to represent treasure locations. They are generally at spots that would require interpolation of the coordinates, so we suspect they are not relevant to the problem. Both "land masses" are completely surrounded by "water", so we aren't sure which is to be considered "its own island". A legend could be helpful.

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K/2 + 9 = 30 What is k

Tcecarenko [31]

K/2+9=30 1. Start by getting rid of the number farthest from K
-9 -9      2. The nines cancel out, 30-9=21
K/2 = 21   3. Next, get the 2 away from K
/-2 /-2 4. The 2 cancel out when divided, 21 divided by -2= -10.5
K = -10.5
K=-10.5

4 0

4 years ago

Find the values of c such that the area of the region bounded by the parabolas

evablogger [386]

For there to be a region bounded by the two parabolas, you first need to find some conditions on $c$ . The two parabolas must intersect each other twice, so you need two solutions to

$16x^2-c^2=c^2-16x^2$

You have

$32x^2=2c^2\implies 16x^2=c^2\implies x=\pm\dfrac{|c|}4$

which means you only need to require that $c\neq0$ . With that, the area of any such bounded region would be given by the integral

$\displaystyle\int_{-4|c|}^{4|c|}\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx$

since $c^2-16x^2>16x^2-c^2$ for all $c\neq0$ . Now,

$\displaystyle\int_{-|c|/4}^{|c|/4}\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx=2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx$

by symmetry across the y-axis. Integrating yields

$\displaystyle2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx=4\int_0^{|c|/4}(c^2-16x^2)\,\mathrm dx$
$=4\left[c^2x-\dfrac{16}3x^3\right]_{x=0}^{x=|c|/4}$
$=c^2|c|-\dfrac{|c|^3}3$
$=\dfrac{2|c|c^2}3=144$
$|c|c^2=216$

Since $216=6^3$ , you have $c=\pm6$ .