Solve this please. |x|<13

Mathematics

1 answer:

Olenka [21]4 years ago

5 0

Answer:

-13 < x < 13

Step-by-step explanation:

Recall that the definition of absolute value is as follows:

$|x|=\left \{ {{x\,,\,\,\mbox{for}\,\,x\geq 0} \atop {-x\,,\,\,\mbox{for}\,\,x$

So we just apply this to solve your inequality:

$|x|$

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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 $c=\pm6$ .