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Tcecarenko [31]
3 years ago
14

Xy and bd are parallel lines

Mathematics
1 answer:
aksik [14]3 years ago
5 0

Answer:

is this a statement or question...?

Step-by-step explanation:

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Factorise x(6x - 5y) - 4(6x - 5y)^2
Fed [463]

Answer:

(6x-5y)(20y-23x)

Step-by-step explanation:

x(6x - 5y) - 4(6x - 5y)^2

= (6x-5y)(x-4(6x-5y))

= (6x-5y)(x-24x+20y)

= (6x-5y)(20y-23x)

5 0
3 years ago
PLEASE i need answer will give brainleist answer!!!
Alik [6]
The company's sales increase by $940
(0.94×1000)
7 0
3 years ago
If 8 identical blackboards are to be divided among 4 schools,how many divisions are possible? How many, if each school mustrecei
MAXImum [283]

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 {11 \choose 3} = 165 . 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 {7 \choose 3} = 35. Thus, there are only 35 ways to distribute the blackboards in this case.

4 0
3 years ago
A mixture contains 40oz of glycol and water, and it is 10% glycol. The mixture is to be strengthened to 25% by adding glycol. Ho
Mazyrski [523]
I found 50, making a rule of three where
 40__10
  x___25

x=10
 40+10= 50
( I hope I'm right, but from what I understand this would be the resolution)



5 0
3 years ago
Solve for x and y<br><br> x-y=11<br><br> 2x+y=19
Ilia_Sergeevich [38]
X=10 y=-1
U can use Photomath for questions like this
7 0
3 years ago
Read 2 more answers
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