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2 years ago

Which of the following graphs model exponential functions? Select all that apply.

Mathematics

1 answer:

Kitty [74]2 years ago

8 0

The graphs model exponential functions will remain positive. Then the correct options are A and D.

<h3>What is an exponent?</h3>

Let a be the initial value and x be the power of the exponent function and b be the increasing factor. The exponent is given as

y = a(b)ˣ

If the initial value a is positive and x is positive, then the graph of the exponent function will approach infinity as x increases.

If the initial value a is positive and x is negative, then the graph of the exponent function will approach zero as x increases.

The graphs model exponential functions will remain positive.

Then the correct options are A and D.

More about the exponent link is given below.

brainly.com/question/5497425

#SPJ1

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GIVING BRAINLIEST, PLS HELP!!!!! :((

blondinia [14]

Step-by-step explanation:

1 Sequence

The numbers are in a right order but starting from 6 and ending at 10....

2 sequence

Numbers are from table of 3 and starting from 3×7 and ending up at 3×10

8 0

3 years ago

<img src="https://tex.z-dn.net/?f=log_%7B8%20x%5E%7B2%7D%20-23x%2B15%7D%282x-2%29%20%5Cleq%200" id="TexFormula1" title="log_{8 x

grandymaker [24]

$\log_{8x^2-23x+15} (2x-2) \leq 0$

The domain:
The number of which the logarithm is taken must be greater than 0.
$2x-2 \ \textgreater \ 0 \\
2x\ \textgreater \ 2 \\
x\ \textgreater \ 1 \\ x \in (1, +\infty)$

The base of the logarithm must be greater than 0 and not equal to 1.
* greater than 0:
$8x^2-23x+15\ \textgreater \ 0 \\ 8x^2-8x-15x+15\ \textgreater \ 0 \\ 8x(x-1)-15(x-1)\ \textgreater \ 0 \\ (8x-15)(x-1)\ \textgreater \ 0 \\ \\ \hbox{the zeros:} \\ 8x-15=0 \ \lor \ x-1=0 \\ 8x=15 \ \lor \ x=1 \\ x=\frac{15}{8} \\ x=1 \frac{7}{8} \\ \\
\hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola op} \hbox{ens upwards} \\
\hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros} \\ \\
x \in (-\infty, 1) \cup (1 \frac{7}{8}, +\infty)$

*not equal to 1:
$8x^2-23x+15 \not= 1 \\
8x^2-23x+14 \not= 0 \\
8x^2-16x-7x+14 \not= 0 \\
8x(x-2)-7(x-2) \not= 0 \\
(8x-7)(x-2) \not= 0 \\
8x-7 \not=0 \ \land \ x-2 \not= 0 \\
8x \not= 7 \ \land \ x \not= 2 \\
x \not= \frac{7}{8} \\ x \notin \{\frac{7}{8}, 2 \}$

Sum up all the domain restrictions:
$x \in (1, +\infty) \ \land \ x \in (-\infty, 1) \cup (1 \frac{7}{8}, +\infty) \ \land \ x \notin \{ \frac{7}{8}, 2 \} \\ \Downarrow \\
x \in (1 \frac{7}{8}, 2) \cup (2, +\infty)
$

The solution:
$\log_{8x^2-23x+15} (2x-2) \leq 0 \\ \\
\overline{\hbox{convert 0 to the logarithm to base } 8x^2-23x+15} \\
\Downarrow \\
\underline{(8x^2-23x+15)^0=1 \hbox{ so } 0=\log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ }
\\ \\
\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1$

Now if the base of the logarithm is less than 1, then you need to flip the sign when solving the inequality. If it's greater than 1, the sign remains the same.

* if the base is less than 1:
$8x^2-23x+15 \ \textless \ 1 \\
8x^2-23x+14 \ \textless \ 0 \\ \\
\hbox{the zeros have already been calculated: they are } x=\frac{7}{8} \hbox{ and } x=2 \\
\hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\
\hbox{the values less than 0 are between the zeros} \\ \\
x \in (\frac{7}{8}, 2) \\ \\
\hbox{including the domain:} \\
x \in (\frac{7}{8}, 2) \ \land \ x \in (1 \frac{7}{8}, 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (1 \frac{7}{8} , 2)$

The inequality:
$\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{flip the sign} \\ 2x-2 \geq 1 \\ 2x \geq 3 \\ x \geq \frac{3}{2} \\ x \geq 1 \frac{1}{2} \\ x \in [1 \frac{1}{2}, +\infty) \\ \\ \hbox{including the condition that the base is less than 1:} \\ x \in [1 \frac{1}{2}, +\infty) \ \land \x \in (1 \frac{7}{8} , 2) \\ \Downarrow \\ x \in (1 \frac{7}{8}, 2)$

* if the base is greater than 1:
$8x^2-23x+15 \ \textgreater \ 1 \\ 8x^2-23x+14 \ \textgreater \ 0 \\ \\ \hbox{the zeros have already been calculated: they are } x=\frac{7}{8} \hbox{ and } x=2 \\ \hbox{the coefficient of } x^2 \hbox{ is greater than 0 so the parabola ope} \hbox{ns upwards} \\ \hbox{the values greater than 0 are between } \pm \infty \hbox{ and the zeros}$

$x \in (-\infty, \frac{7}{8}) \cup (2, +\infty) \\ \\ \hbox{including the domain:} \\ x \in (-\infty, \frac{7}{8}) \cup (2, +\infty) \ \land \ x \in (1 \frac{7}{8}, 2) \cup (2, +\infty) \\ \Downarrow \\ x \in (2, \infty)$

The inequality:
$\log_{8x^2-23x+15} (2x-2) \leq \log_{8x^2-23x+15} 1 \ \ \ \ \ \ \ |\hbox{the sign remains the same} \\ 2x-2 \leq 1 \\ 2x \leq 3 \\ x \leq \frac{3}{2} \\ x \leq 1 \frac{1}{2} \\ x \in (-\infty, 1 \frac{1}{2}] \\ \\ \hbox{including the condition that the base is greater than 1:} \\ x \in (-\infty, 1 \frac{1}{2}] \ \land \ x \in (2, \infty) \\ \Downarrow \\ x \in \emptyset$

Sum up both solutions:
$x \in (1 \frac{7}{8}, 2) \ \lor \ x \in \emptyset \\ \Downarrow \\
x \in (1 \frac{7}{8}, 2)$

The final answer is:
$\boxed{x \in (1 \frac{7}{8}, 2)}$

5 0

3 years ago

Solve for Z.<br> 3z - 12 =48

marysya [2.9K]

Answer:

Z = 20

Step-by-step explanation:

3 x 20 = 60

60 - 12 = 48

4 0

3 years ago

Read 2 more answers

Bryce is looking at a map of a theme park. The map is laid out in a coordinate system. Bryce is at (2, 3). The roller coaster is

Sunny_sXe [5.5K]

Answer:

Bryce is closer to the <u>roller coaster</u> as compared to the water ride.

Step-by-step explanation:

Given :

Bryce is at point (2,3)

Roller coaster is at point (7,8)

Water ride is at point (9,1)

To find whether Bryce is closer to roller coaster or the water ride.

Solution.

In order to determine the closeness of Bryce from either of the two rides, we will need to find the distance of Bryce from each of them. We can apply distance formula in order to do so.

By distance formula to find distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ .

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Distance between Bryce (2,3) and roller coaster (7,8) can be given as:

$d_1=\sqrt{(7-2)^2+(8-3)^2}$

$d_1=\sqrt{(5)^2+(5)^2}$

$d_1=\sqrt{25+25}$

$d_1=\sqrt{50}$

$d_1=7.07$ units

Distance between Bryce (2,3) and water ride (9,1) can be given as:

$d_2=\sqrt{(9-2)^2+(1-3)^2}$

$d_2=\sqrt{(7)^2+(-2)^2}$

$d_2=\sqrt{49+4}$

$d_2=\sqrt{53}$

$d_2=7.28$ units

Thus, we can see that Bryce is closer to the roller coaster as compared to the water ride.

3 0

3 years ago

Need help ( 8th grade math)

kifflom [539]

It would be -3/1. This is because you find where it crosses the Y axis which is at 4. Then you go down three and over one to give you the next even point. Which going down to the right means it’s negative. Hope this helps! Please mark branlist if possible!!

6 0

3 years ago

Read 2 more answers