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3 years ago

SELECT ALL of the Angles that are supplementary to Angle 7 NO LINKS

Mathematics

1 answer:

sergejj [24]3 years ago

5 0

Answer:

The angles are 6,4,8and2 hope this helps

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3 years ago

Read 2 more answers

State the Pythagorean Theorem in your own words

Solnce55 [7]

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A ^2 + B^2 =C^2. The Pythagorean Theorem is a statement about triangles containing a right angle

Step-by-step explanation:

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4 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

The ratio of children to adults at the football game was 23. If there were 140 children at the football game,

inessss [21]

Answer:

Therefore, the number of adults at the football game was 6.

Step-by-step explanation:

Let A represents the number of adults and C represents the number of children.

Therefore, the ratio of children to adults is given as follows:

23:1 = C:A ....................... (1)

Since C = 140, this is substituted into equation (1) as follows:

23:1 = = 140:A ....................... (2)

Transforming equation (2) and solve for A, we have:

23 / (23 + 1) = 140 / (140 + A)

23 / 24 = 140 / (140 + A)

0.958333333333333 = 140 / (140 + A)

0.958333333333333 (140 + A) = 140

(0.958333333333333 * 140) + 0.958333333333333A = 140

134.166666666667 + 0.958333333333333A = 140

0.958333333333333A = 140 -134.166666666667

0.958333333333333A = 5.833333333333

A = 5.833333333333 / 0.958333333333333

A = 6.08695652173878

Rounding to a whole number, we have:

A = 6

Therefore, the number of adults at the football game was 6.

5 0

3 years ago

SELECT ALL of the Angles that are supplementary to Angle 7 **NO LINKS**

SELECT ALL of the Angles that are supplementary to Angle 7 NO LINKS