Ask question

Ask question

All categories

3 years ago

13

A school store ordered 12 boxes of notebooks. For each box they paid for, their bank account said -$16.50. What did their bank a

ccount show after they paid for all 12 boxes? (Write your answer in dollars $)

1 answer:

SashulF [63]3 years ago

6 0

Answer:

-198 dollars

Step-by-step explanation:

You might be interested in

<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

What is the coefficient constant and factor of 5x^2 3x 12

Yanka [14]

<span>Factoring 5x2</span> + 3x - 12. The first term is, <span>5x2</span> its coefficient is 5 . The middle term is, +3x its coefficient is 3 . The last term, "the constant", is -<span>12 heres your answer your welcome</span>

3 0

3 years ago

Show two different ways 3/4×4

agasfer [191]

Answer:

0.75 x 4 and 75 x 0.04, hope this helps.

Step-by-step explanation:

6 0

4 years ago

Read 2 more answers

Can anyone help me ?

nydimaria [60]

Which TWO expressions are equivalent to<em> </em><em>(x - y) </em>2/3 - 3/4 <em>(y-x)</em>

8 0

3 years ago

Missing side of the triangle<br> 5 yds<br> 4yds

Vitek1552 [10]

Answer:

6.4yds if you are finding the hyp

Step-by-step explanation:

5^2+4^2=square root41

square root of 41 is 6.4

4 0

3 years ago