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Andreas93 [3]
4 years ago
11

If anyone is good in Pythagoras theoram please help me

Mathematics
2 answers:
Yanka [14]4 years ago
4 0

Answer:

a^2+B^2=c^2

Step-by-step explanation:

disa [49]4 years ago
3 0

Step-by-step explanation:

(Hyp)² = (Base)² + (Height)²

C² = a² + b²

C= √a²+b²

This is the formula to find Hypotenuse of a right angled triangle

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What would I do for distributive property with this expression?
vovikov84 [41]
You can use the box method and multiply it out like in this pic

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3 years ago
On a coordinate map of the school cultural festival, the information booth is located at (4,5). The ticket booth is located 4 un
monitta

Answer:

B.

Step-by-step explanation:

6 0
3 years ago
An event manager recorded the number of people in different age groups who attended a music concert: A histogram titled Concert
OverLord2011 [107]

Answer:

Age Group Number of People 18–24 80 25–31 120 32–38 40 39–45 20

Step-by-step explanation:

The data in each bar is how many people in that group.

So 18-24 has 80 people

25-31 has 120 people

32 -38 has 40 people

39-45 has 20 people

4 0
4 years ago
1700 + 2029.8x = 9500 solve for x
Novosadov [1.4K]

Step-by-step explanation:

1700 + 2029.8x = 9500

2029.8x=9500-1700

x=7800/2029.8

x=3.85

6 0
3 years ago
Read 2 more answers
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
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