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3 years ago

PLEASE HELP ME YALL I NEED YOUR HELP LIKE A LOT. PLEASEEEE I DONT UNDERSTAND NOTHING I AM VERY DUMB

Mathematics

1 answer:

astraxan [27]3 years ago

5 0

(>_ means greater than or equal to, and <_ means less than or equal to)

Answer:

9: Domain: 0<_x<_10

10: Range: 0<_y<_32

11: Domain: 0<_x<_∞ Range: 10<_y<_∞

12: Domain: 0<_x<_∞ Range: y=25

13: Domain: 0<_x<_15 Range: 0<_y<_15

Step-by-step explanation:

Just look at the highest and lowest points for the x axis and the y axis, and that's your domain and range respectively

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[HELP] For j(x) = <img src="https://tex.z-dn.net/?f=5%5Ex%5E-%5E3" id="TexFormula1" title="5^x^-^3" alt="5^x^-^3" align="absmidd

Alina [70]

The difference quotient of the function that has been presented to us will turn out to be 5.

<h3>How can I calculate the quotient of differences?</h3>

In this step, we wish to determine the difference quotient for the function that was supplied.

To begin, keep in mind that the difference quotient may be calculated by:

Lim h->0 $\frac{f(x+h)-f(x)}{h}$

Now, for the purpose of the function, we need this:

Then we will have:

$\begin{aligned}&\lim _{h \rightarrow 0} \frac{j(x+h)-j(x)}{h} \\&\lim _{h \rightarrow 0} \frac{5 *(x+h)-3-5 * x+3}{h} \\&\lim _{h \rightarrow 0} \frac{5 x+5 h-3-5 x+3}{h} \\&\lim _{h \rightarrow 0} \frac{5 h}{h}=5\end{aligned}$

j(x) = 5x - 3

Then the following will be true:

Therefore, 5 is the value of the difference quotient for j(x) is %

Read the following if you are interested in finding out more about difference quotients:

brainly.com/question/15166834

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6 0

1 year ago

Suppose you have two urns with poker chips in them. Urn I contains two red chips and four white chips. Urn II contains three red

Neporo4naja [7]

Answer:

Multiple answers

Step-by-step explanation:

The original urns have:

Urn 1 = 2 red + 4 white = 6 chips
Urn 2 = 3 red + 1 white = 4 chips

We take one chip from the first urn, so we have:

The probability of take a red one is : $\frac{1}{3}$ (2 red from 6 chips(2/6=1/2))

For a white one is: $\frac{2}{3}$ (4 white from 6 chips(4/6=(2/3))

Then we put this chip into the second urn:

We have two possible cases:

First if the chip we got from the first urn was white. The urn 2 now has 3 red + 2 whites = 5 chips
Second if the chip we got from the first urn was red. The urn two now has 4 red + 1 white = 5 chips

If we select a chip from the urn two:

In the first case the probability of taking a white one is of: $\frac{2}{5}$ = 40% ( 2 whites of 5 chips)
In the second case the probability of taking a white one is of: $\frac{1}{5}$ = 20% ( 1 whites of 5 chips)

This problem is a dependent event because the final result depends of the first chip we got from the urn 1.

For the fist case we multiply :

$\frac{4}{6}$ x $\frac{2}{5}$ = $\frac{4}{15}$ = 26.66% ( $\frac{4}{6}$ the probability of taking a white chip from the urn 1, $\frac{2}{5}$ the probability of taking a white chip from urn two)

For the second case we multiply:

$\frac{1}{3}$ x $\frac{1}{5}$ = $\frac{1}{30}$ = .06% ( $\frac{1}{3}$ the probability of taking a red chip from the urn 1, $\frac{1}{5}$ the probability of taking a white chip from the urn two)