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4 years ago

For number 11a - 11d select true pg false for each sentence

Mathematics

1 answer:

Katarina [22]4 years ago

6 0

What are the choices

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<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

Question on picture

fenix001 [56]

I think you have the right one.

6 0

3 years ago

A cylinder. The diameter of the base is 5 feet and the height of the cylinder is 12 feet.

MaRussiya [10]

Answer:

235.5 ft³

Step-by-step explanation:

V = Ab×h

= πr²×h

= π(2.5)²×12

= 3.14×6.25×12

= 235.5 ft³

6 0

3 years ago

The sum of two numbers is 24. The second number is 4 more than the first number. Let a represent the first number

OverLord2011 [107]

Answer:

a=10, b=14

Step-by-step explanation:

a+b = 24

b=a+4

a+(a+4)=24

2a+4=24

2a=20

a=10

b=14

6 0

3 years ago

Sofia and Tess will each randomly choose one of the 10 integers from 1 to 10. What is the probability that neither integer chose

Ann [662]

There are $10^2=100$ possible pairs they can choose. Let $S$ and $T$ denote the integers that Sofia and Tess, respectively, choose, so that $(S,T)$ denotes the integer pair.

The event that either $S=T^2$ or $S^2=T$ has 5 possible ways of occurring:

$\{(1,1),(2,4),(4,2),(3,9),(9,3)\}$

So the probability of this happening is $\dfrac5{100}=\dfrac1{20}=0.05=5\%$ . You're interested in the complement, which means the answer is 95%.

3 0

3 years ago