4k – 6 = -2k – 16

There are 6 diagonals that can be drawn from one vertex of an octagon, true or false.

FinnZ [79.3K]

It's False; an octagon has 8 vertices. When you remove the starting vertex and the two adjacent vertices we're left with 5 possible diagonals

6 0

4 years ago

What's the value? A.-20 B.-4 C.4 D.20

ale4655 [162]

Answer:

Option B is the correct option.

Step-by-step explanation:

${(4 - 2)}^{3} - 3 \times 4$

Subtract the numbers

$= {(2)}^{3 } - 3 \times 4$

Multiply the numbers

$= {(2)}^{3} - 12$

Evaluate the power

$= 8 - 12$

Calculate the difference

$= - 4$

Hope this helps..

Best regards!!

4 0

3 years ago

Read 2 more answers

Find the solution set of this equality |2x+4|>6

lisov135 [29]

The answer to the question is 1.

4 0

3 years ago

What percent of the first 20 natural numbers are prime numbers?

Brilliant_brown [7]

The primes are 2, 3, 5, 7, 11, 13, 17, 19. There are 8 of them, so they make up 40% of the numbers 1–20.

3 0

3 years ago

Read 2 more answers

The points A(1, 4), B(5,1) lie on a circle. The line segment AB is a chord. Find the equation of a diameter of the circle.

tangare [24]

Check the picture below.

well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.

bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.

$\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}$

so, it passes through the midpoint of AB,

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{5+1}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(3~~,~~\cfrac{5}{2} \right)$

so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)

$\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{\frac{5}{2}}) \stackrel{slope}{m}\implies \cfrac{4}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{\cfrac{5}{2}}=\stackrel{m}{\cfrac{4}{3}}(x-\stackrel{x_1}{3})\implies y-\cfrac{5}{2}=\cfrac{4}{3}x-4 \\\\\\ y=\cfrac{4}{3}x-4+\cfrac{5}{2}\implies y=\cfrac{4}{3}x-\cfrac{3}{2}$

4 0

3 years ago

4k – 6 = -2k – 16 – 2