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3 years ago

Find the distance and midpoint for the points (5,5) and (1,2).

Mathematics

1 answer:

irinina [24]3 years ago

4 0

Answer:

try and subtract the orgin is 0,0 so maybe it will help

Step-by-step explanation:

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Simplify the expression (2-5i)-(4-4i)

RUDIKE [14]

(2 - 5i) - (4 - 4i)
-2 - i is your answer. Hope it helps! :) If you could vote my answer as the brainiest, that would be awesome! :)

3 0

3 years ago

HELPPPPPPP MEEEEEEEEEEEEE

Evgesh-ka [11]

Number 1 is 2 because first you find out how many times 42 goes into 84 so you do 42x1, 42x2 and so on until you can't do anymore without going over 84 so 42x2 is 84 exactly so you put the two on top of the "house" above the 4 and put 84 under 84 then subtract 84-84=0 so you don't continue dividing and the number on top of the "house" is the answer

3 0

4 years ago

Please help me with this?

makvit [3.9K]

Number 2 is 25 that fors sure but i dont wanna give you the 3rd i dont know why bt i dont :/

8 0

4 years ago

Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p

posledela

Answer:

P=0.0000037

P=0.00037%

Step-by-step explanation:

Probability

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

$\displaystyle \frac{4}{52}=\frac{1}{13}$

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

$\displaystyle \frac{3}{51}=\frac{1}{17}$

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

$\displaystyle \frac{2}{50}=\frac{1}{25}$

Finally, the probability of getting the remaining ace out of the 49 cards is:

$\displaystyle \frac{1}{49}$

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

$\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}$

$\displaystyle P= \frac{1}{270,725}$

P=0.0000037

P=0.00037%

3 0

3 years ago

What is the eighth term of the geometric sequence 6, 18, 54

Rainbow [258]

Answer:

= 4374.

Step-by-step explanation:

it is important to understand the pattern hidden in such problem.Let’s give it a try 2, 6, 18, 54 and so on.It can be written as 2, 3*2, 9*2, 27*2 and so on.

This can be further written as2 (1, 3, 9, 27, and so on) as 2 is common in every term.Now if you see the chain 1,3,9, 27 and so….you will see a pattern hidden i.e. 3=1*3 ,9=1*3*3, 27=1*3*3*3 now 27 is the 4th term consist of three 3. So 8th term would consist of seven 3. 8th would be 8th term = 1*3*3*3*3*3*3*3 = 2187 Hence the 8th term for the series 2,6,18,54 would be= 2*2187 = 4374.

4 0

3 years ago