Which does not show a direct variation between x and y?

2 answers:

8 0

A direct variation is a mathematical relationship between two variables that can be expressed by an equation in which one variable is equal to a constant times the other. In other words, a direct variation is where $y = x * (constant)$ .

In answer a, $y = x*.5$ , so it is a direct variation.
In answer c, $y = x* \frac{1}{9}$ , so it is a direct variation.
In answer d, $y = x*2$ , so it is a direct variation,

Only answer b is left, which means the answer must be 'b'.

We also know 'b' is the answer because it cannot be expressed as <span> $y = x * (constant)$ . Instead, it is expressed as </span> $y = \frac{constant}{x}$ , which is not the same thing and is therefore not a direct variation.

Hope I helped, and let me know if you have any questions :)

Margarita [4]3 years ago

4 0

<span>A relationship between two variables in which one is a constant multiple of the other. In particular, when one variable changes the other changes in proportion to the first. If b is directly proportional to a, the equation is of the form b = ka (where k is a constant).</span>
B.y = 9/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 $c=\pm6$ .