X=-6
Hope it helps!!
The answer is -6!!
<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>
The domain of the function given above is the value of w that would make it reasonable. Among the numbers, w cannot be zero because that would make the fraction 70/0 undefined. So the domain of the function is from negative infinity to positive infinity but not zero.
Answer:
x = 100√3
Step-by-step explanation:
tan(30º) = x/300
x = 300 tan(30º)
x = 300/√3
x = 100√3
(2,-2)(4,10)
slope = (10 - (-2) / (4 - 2) = (10 + 2) / 2 = 12/2 = 6
The slope of the original equation is 6, but we are looking for a perpendicular line, so we need the negative reciprocal slope. All that means is " flip " the slope and change the sign. 6 or 6/1.....flip it...1/6...change the sign....-1/6. So we will need the slope of -1/6.
y = mx + b
slope(m) = -1/6
(2,1)...x = 2 and y = 1
now we sub and find b, the y intercept
1 = -1/6(2) + b
1 = -1/3 + b
1 + 1/3 = b
3/3 + 1/3 = b
4/3 = b
so ur perpendicular equation is : y = -1/6x + 4/3
