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2 years ago

Which function represents a reflection of f(x) = 5(0.8)x across the x-axis?

Mathematics

1 answer:

lukranit [14]2 years ago

8 0

The function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

<h3>Reflection of functions and coordinates</h3>

Images that are reflected are mirror images of each other. When a point is reflected across the line y = x, the x-coordinates and y-coordinates change their position. In a similar manner, when a point is reflected across the line y = -x, the coordinates <u>changes position but are negated.</u>

Given the exponential function below

f(x) = 5(0.8)^x

If the function f(x) is reflected over the x-axis, the resulting function will be

-f(x)

This means that we are going to negate the function f(x) as shown;

f(x) = -5(0.8)^x

Hence the function represents a reflection of f(x) = 5(0.8)x across the x-axis is f(x) = -5(0.8)^x

Learn more on reflection here: brainly.com/question/1908648

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Given f(x)=x^2+5 and g(x)=2x^3-1 find (f/g)(-3)

AlekseyPX

For this case we have the following functions:

$f (x) = x ^ 2 + 5\\g (x) = 2x ^ 3-1$

By definition of composition of functions we have:

$(f / g) (x) = \frac {f (x)} {g (x)}$

Then substituting:

$f (-3) = (- 3) ^ 2 + 5 = 9 + 5 = 14\\g (-3) = 2 (-3) ^ 3-1 = 2 (-27) -1 = -54-1 = -55$

So:

$(f / g) (- 3) = \frac {14} {- 55} = - 0.25$

Answer:

$(f / g) (- 3) = \frac {14} {- 55} = - 0.25$

7 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2