Answer:
160
Step-by-step explanation:
13² - 3² = 13 × 13 - 3 × 3 = 169 - 9 = 160
Answer:
There are 165 ways to distribute the blackboards between the schools. If at least 1 blackboard goes to each school, then we only have 35 ways.
Step-by-step explanation:
Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.
The problem reduces to take 11 consecutive spots which we will use to localize the balls and the sticks and select 3 places to put the sticks. The amount of ways to do this is
As a result, we have 165 ways to distribute the blackboards.
If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is
. Thus, there are only 35 ways to distribute the blackboards in this case.
Answer:
I believe the second one is d
4x-2 + 7x-18 + 6x+6 = 360 degrees
combine like terms
17x-14 = 360
add 14 to both sides
17x =374
divide both sides by 17
x = 374 / 17 = 22
JL = 7x-18
replace x with 22 and calculate
7(22) - 18 = 154 -18 = 136 degrees
Answer:
A -60i - 14j
Step-by-step explanation:
u = -9i + 8j
v = 7i + 5j
2u = 2(-9i + 8j) = -18i + 16j
6v = 6(7i + 5j) = 42i + 30j
2u - 6v
(-18i + 16j) - (42i + 30j)
(-18i - 42i) + (16j-30j)
-60i - 14j