Answer:
y = a + b x + c .
Step-by-step explanation:
Also (How to do it!):
We can find the vertex of a quadratic equation with these steps:
Get the equation in the form y = ax2 + bx + c.
Calculate -b / 2a. This is the x-coordinate of the vertex.
To find the y-coordinate of the vertex, simply plug the value of -b / 2a into the equation for x and solve for y.
The way we would answer the question would be to express it as
100-9(n-1) with n being the nth term of the pattern.
Why is this? We know we subtract 9 each time and START from 100.
Remember though that 100 is the first term in the sequence. So, we would have (n-1)*9 to subtract 0 from our 1st term and 9 from our second term.
Also, we have 9*(n-1) because we are just subtracting multiples of 9. It is always going to subtract by 9.
If you needed the answer here it is...
The answer to the problem would be
100-9(13-1)
= 100-9(12)
= 100-108
= -8
<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>
Answer:
10 and 11
Step-by-step explanation:
Answer:
i cant see the photo, but i would love to help
Step-by-step explanation: