Answer:
Step-by-step explanation:
roots of a complex number is given by DeMoivre's formula.
![\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}](https://tex.z-dn.net/?f=%5Csf%20%5Cboxed%7B%5Cbf%20r%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%5Cleft%5BCos%20%5Cdfrac%7B%5Ctheta%20%2B%202%5Cpi%20k%7D%7Bn%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Ctheta%2B2%5Cpi%20k%7D%7Bn%7D%5Cright%5D%7D)
Here, k lies between 0 and (n -1) ; n is the exponent.
a = -1 and b = √3
n = 4
For k = 0,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%20%2B0%7D%7B4%7D%2BiSin%20%20%5C%20%5Cdfrac%7B%5Cdfrac%7B-%5Cpi%7D%7B3%7D%2B0%7D%7B4%7D%5Cright%5D%20%5C%5C%5C%5C%5C%5Cz%3D%20%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%5BCos%20%5C%20%5Cdfrac%7B%20-%5Cpi%20%20%7D%7B12%7D%2BiSin%20%20%5C%20%5Cdfrac%7B-%5Cpi%7D%7B12%7D%5Cright%5D%5C%5C%5C%5C%5C%5Cz%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5B-Cos%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D-i%20%5C%20Sin%20%5C%20%5Cdfrac%7B%5Cpi%7D%7B12%7D%5Cright%5D)
For k =1,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B5%5Cpi%7D%7B12%7D%5Cright%5D)
For k =2,
![z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B11%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 3,
![\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B17%5Cpi%7D%7B12%7D%5Cright%5D)
For k = 4,
![\sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]](https://tex.z-dn.net/?f=%5Csf%20z%20%3D%5Csqrt%5B4%5D%7B10%7D%5Cleft%5BCos%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%2Bi%20%5C%20Sin%20%5C%20%5Cdfrac%7B23%5Cpi%7D%7B12%7D%5Cright%5D)
Answer:
-6x + 15 = - 69 which x = 14
Step-by-step explanation:
You cannot solve this equation. There is no equal sign anywhere. Either you mean
-6x = 15 in which case the answer is
-6x/-6 = 15/-6
x = - 2 1/2 or x= - 2.5
or you mean that you use the distributive property
- 3(2x + 5)
You may have left something out of this.
-6x + 15 = - 69
You can do it now that you have the complete equation. But you need the entire equation before you can answer the question.
Subtract - 15 from both sides.
-6x + 15 - 15 = -69 - 15
-6x = - 84
Divide by - 6
-6x/-6 = - 84/-6
x = 14
But what you want is
-6x + 15
-6*14 + 15
-84 + 15
-69
-6x + 15 = - 69
when x = 14 which is usually what you are asked.
it should be 100/3 or 33.3 repeating
<span>There are several ways to do this problem. One of them is to realize that there's only 14 possible calendars for any year (a year may start on any of 7 days, and a year may be either a leap year, or a non-leap year. So 7*2 = 14 possible calendars for any year). And since there's only 14 different possibilities, it's quite easy to perform an exhaustive search to prove that any year has between 1 and 3 Friday the 13ths.
Let's first deal with non-leap years. Initially, I'll determine what day of the week the 13th falls for each month for a year that starts on Sunday.
Jan - Friday
Feb - Monday
Mar - Monday
Apr - Thursday
May - Saturday
Jun - Tuesday
Jul - Thursday
Aug - Sunday
Sep - Wednesday
Oct - Friday
Nov - Monday
Dec - Wednesday
Now let's count how many times for each weekday, the 13th falls there.
Sunday - 1
Monday - 3
Tuesday - 1
Wednesday - 2
Thursday - 2
Friday - 2
Saturday - 1
The key thing to notice is that there is that the number of times the 13th falls upon a weekday is always in the range of 1 to 3 days. And if the non-leap year were to start on any other day of the week, the numbers would simply rotate to the next days. The above list is generated for a year where January 1st falls on a Sunday. If instead it were to fall on a Monday, then the value above for Sunday would be the value for Monday. The value above for Monday would be the value for Tuesday, etc.
So we've handled all possible non-leap years. Let's do that again for a leap year starting on a Sunday. We get:
Jan - Friday
Feb - Monday
Mar - Tuesday
Apr - Friday
May - Sunday
Jun - Wednesday
Jul - Friday
Aug - Monday
Sep - Thursday
Oct - Saturday
Nov - Tuesday
Dec - Thursday
And the weekday totals are:
Sunday - 1
Monday - 2
Tuesday - 2
Wednesday - 1
Thursday - 2
Friday - 3
Saturday - 1
And once again, for every weekday, the total is between 1 and 3. And the same argument applies for every leap year.
And since we've covered both leap and non-leap years. Then we've demonstrated that for every possible year, Friday the 13th will happen at least once, and no more than 3 times.</span>
