A paper bag has 3 red and 7 blue marbles. What is the probability of getting f getting a blue marble on the first try

Mathematics

1 answer:

solong [7]3 years ago

5 0

Answer: 7/10

Step-by-step explanation:

First you add the total number of marbles. Then, you take the total number of blue marbles, and make that your denominator. The total number of marbles is your numerator.

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The distance $l$ between $(x, y)$ and $(3, 0)$ is:
$l=\sqrt{(0-y)^{2}+(3-x)^{2}}=\sqrt{y^{2}+(3-x)^{2}}$

Since the equation of the hyperbola is $xy=8$ , we can get $y$ by itself and end up with
$y=\frac{8}{x}$

which we can plug into our distance formula:
$l=\sqrt{(\frac{8}{x})^{2}+(3-x)^{2}}$

To make calculation easier, we'll square both sides:
$l^{2}=(\frac{8}{x})^2+(3-x)^{2}$
and create a new variable $m=l^{2}$ :
$m=(\frac{8}{x})^2+(3-x)^{2}$
$\rightarrow m=\frac{64}{x^{2}}+9-6x+x^{2}$

Differentiate both sides:
$\frac{dm}{dx}=-\frac{128}{x^{3}}-6+2x$
Minimum distance is achieved when $\frac{dm}{dx}=0$ :
$-\frac{128}{x^{3}}-6+2x=0$
$\rightarrow -128-6x^{3}+2x^{4}=0$
$\rightarrow 2(x^{4}-3x^{3}-64)=0$
$\rightarrow x^{4}-3x^{3}-64=0$

To find a value of $x$ , you can use methods like synthetic division and get the answer $x=4$

Plug into $xy=8$ :
$4y=8$
$\rightarrow y=2$

So the closest point on the hyperbola to $(3, 0)$ is $(4, 2)$