(6x - 5)(6x - 5)(6x - 5)(6x - 5)(6x

Some people think it is unlucky if the 13th day of month falls on a Friday. show that in that there year (non-leap or leap) ther

Vlad1618 [11]

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.

5 0

2 years ago

If you have a cubic polynomial of the form y = ax^3 + bx^2 + cx + d and lets say it passes through the points (2,28), (-1, -5),

nikklg [1K]

Step-by-step explanation:

Step 1: Solve using the first point

(2, 28)

$28 = a(2)^3 + b(2)^2 + c(2) + d$

$28 = 8a + 4b + 2c + d$

Step 2: Solve using the second point

(-1, -5)

$-5 = a(-1)^3 + b(-1)^2 + c(-1) + d$

$-5 = -a + b - c + d$

Step 3: Solve using the third point

(4, 220)

$220 = a(4)^3 + b(4)^2 + c(4) + d$

$220 = 64a + 16b + 4c + d$

Step 4: Solve using the fourth point

(-2, -20)

$-20 = a(-2)^3 + b(-2)^2 + c(-2) + d$

$-20 = -8a + 4b - 2c + d$

Step 5: Combine the first and fourth equations

$28 - 20 = 8a - 8a + 4b + 4b + 2c - 2c + d + d$

$8 = 8b + 2d$

$8 - 8b = 8b - 8b + 2d$

$(8 -8b)/2 = 2d/2$

$4 - 4b = d$

Step 6: Solve for c in the second equation

$-5 + 5 = -a + b - c + d + 5$

$0 + c = -a + b - c + c + d + 5$

$c = -a + b + d + 5$

Step 7: Substitute d with the stuff we got in step 5

$c = -a + b + (4 - 4b) + 5$

$c = -a + b + 4 - 4b + 5$

$c = -a - 3b + 9$

Step 8: Substitute d and c into the first equation

$28 = 8a + 4b + 2(-a - 3b + 9) + (4 - 4b)$

$28 = 8a + 4b - 2a - 6b + 18 + 4 - 4b$

$28 - 22 = 6a - 6b + 22 - 22$

$6 / 6 = (6a - 6b) / 6$

$1 + b = a - b + b$

$1 + b = a$

Step 9: Substitute a, b, and c into the third equation

$220 = 64(1 + b) + 16b + 4(-(1 + b) - 3b + 9) + (4 - 4b)$

$220 = 64 + 64b + 16b + 4(-1 - b - 3b + 9) + 4 - 4b$

$220 - 100 = 60b + 100 - 100$

$120 / 60 = 60b / 60$

$2 = b$

Step 10: Find a using b = 2

$a = b + 1$

$a = (2) + 1$

$a = 3$

Step 11: Find c using a = 3 and b = 2

$c = -a - 3b + 9$

$c = -(3) - 3(2) + 9$

$c = -3 - 6 + 9$

$c = 0$

Step 12: Find d using b = 2

$d = 4 - 4b$

$d = 4 - 4(2)$

$d = 4 - 8$

$d = -4$

Answer: $a = 3, b = 2, c = 0,d = -4$

6 0

2 years ago

Read 2 more answers

He is paid a commission of 9 percent of his first $6,000 in sales during the month and 14 percent on all sales over $6,000. What

Agata [3.3K]

Answer:

$1,956.80

Step-by-step explanation:

For amounts over $6000, the commission can be computed as ...

0.14s -300 . . . . . . for sales (s) ≥ 6000

So, for $16,120 in sales, the commission is ...

0.14×$16,120 -300 = $2,256.80 -300 = $1,956.80

__

The commission schedule suggests that for larger amounts, you divide the problem into two parts: calculate the commission on $6000, and separately calculate the commission on the amount over $6000.

0.14(s -6000) + 0.09(6000)

= 0.14s - 0.14·6000 +0.09·6000

= 0.14s -300 . . . . the formula used above for s ≥ 6000

6 0

2 years ago

Please help, answer and show me the work. Thanks also solve for x

vova2212 [387]

3. x+100 = 90 alternate interior angles are equal

subtract 100 from each side

x+100 - 100 = 90-100

x = -10

4. 17x+1 = 120 alternate exterior angles are equal

subtract 1 from each side

17x+ 1 -1 = 120 -1

17x = 119

divide each side by 17

17x/17 = 119

x =7

4 0

2 years ago

(5,1); m=2 write an equation of the line

gtnhenbr [62]

Hey there! :)

To find an equation of a line that passes through (5, 1) and has a slope of 2, we'll need to plug our known variables into the slope-intercept equation.

Slope-intercept equation : y = mx + b ; where m=slope, b=y-intercept

Since we're already given the slope, all we really need to do is find the y-intercept.

We can do this by plugging our known values into the slope-intercept equation.

y = mx + b

Since we're trying to find "b," we need to plug in "y, m, x" into our formula.

(1) = (2)(5) + b

Simplify.

1 = 10 + b

Subtract 10 from both sides.

1 - 10 = b

Simplify.

-9 = b

So, our y-intercept is 9!

Now, we can very simply plug our known values into slope-intercept form.

y = mx + b

y = 2x - 9 → final answer

~Hope I helped!~

3 0

2 years ago

(6x - 5)(6x - 5)(6x - 5)(6x - 5)(6x - 5) = 0