Answer:a=-4,000
Step-by-step explanation:
Answer:
p = - 5, p = - 1
Step-by-step explanation:
Given
- 
= 
Multiply through by the LCM of (p - 5)(p + 3)
3p(p + 3) - 2(p - 5) = p(p - 5) ← distribute parenthesis
3p² + 9p - 2p + 10 = p² - 5p
3p² + 7p + 10 = p² - 5p ( subtract p² - 5p from both sides )
2p² + 12p + 10 = 0 ← divide through by 2
p² + 6p + 5 = 0 ← in standard form
(p + 1)(p + 5) = 0 ← in factored form
Equate each factor to zero and solve for p
p + 1 = 0 ⇒ p = - 1
p + 5 = 0 ⇒ p = - 5
1) Find the value of x and y.
2) Then use the Order of Operations to help you out with the computation.
Answer:
The answer is -3.4. If you have to round it, round it down to 3.
Step-by-step explanation:
<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>
