well, in short, the students and chaperones are at a 2 : 9 ratio, and we know there are 171 students, how many chaperones then, so long they maintain a 2:9 ratio?
If a regular polygon has 40 sides the measure of each interior angle will be
2 answers:
The formula for interior angles is:
Using this, plug in 40 as <em>n </em>and find <em>s</em>.
If you do that, you will get <em>s=6840.
</em>Now that you have the <u>sum</u> of interior angles, you have to find one single measure of an angle in your regular polygon.
<em>If 40 equal sides in a regular polygon results to 6,840°, then one vertex should be the sum of the interior angles divided by the total number of sides.
</em>
In other words, you will have to find 6840°/40. You will get 171°. That means that the measure of a single interior angle is 171*.
Using this, plug in 40 as <em>n </em>and find <em>s</em>.
If you do that, you will get <em>s=6840.
</em>Now that you have the <u>sum</u> of interior angles, you have to find one single measure of an angle in your regular polygon.
<em>If 40 equal sides in a regular polygon results to 6,840°, then one vertex should be the sum of the interior angles divided by the total number of sides.
</em>
In other words, you will have to find 6840°/40. You will get 171°. That means that the measure of a single interior angle is 171*.
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Answer:
Step-by-step explanation:
Hello,
degree 5
4 is a zero of multiplicity 3 -> (x-4)^3 is a factor
-4 is the only other zero, so the multiplicity is 5-3=2 -> (x+4)^2 is a factor
leading coefficient is 4 so we can write
If there is something that you do not understand or you are blocked somewhere let us know what / where.
Thank you.
Answer:
1) (x + 3)(3x + 2)
2) x= +/-root6 - 1 by 5
Step-by-step explanation:
3x^2 + 11x + 6 = 0 (mid-term break)
using mid-term break
3x^2 + 9x + 2x + 6 = 0
factor out 3x from first pair and +2 from the second pair
3x(x + 3) + 2(x + 3)
factor out x+3
(x + 3)(3x + 2)
5x^2 + 2x = 1 (completing squares)
rearrange the equation
5x^2 + 2x - 1 = 0
divide both sides by 5 to cancel out the 5 of first term
5x^2/5 + 2x/5 - 1/5 = 0/5
x^2 + 2x/5 - 1/5 = 0
rearranging the equation to gain a+b=c form
x^2 + 2x/5 = 1/5
adding (1/5)^2 on both sides
x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2
(x + 1/5)^2 = 1/5 + 1/25
(x + 1/5)^2 = 5 + 1 by 25
(x + 1/5)^2 = 6/25
taking square root on both sides
root(x + 1/5)^2 = +/- root(6/25)
x + 1/5 = +/- root6 /5
shifting 1/5 on the other side
x = +/- root6 /5 - 1/5
x = +/- root6 - 1 by 5
x = + root6 - 1 by 5 or x= - root6 - 1 by 5
