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Marianna [84]
2 years ago
9

Using the digits 0 to 9, without repetition, fill the blanks.

Mathematics
1 answer:
Alexandra [31]2 years ago
4 0
First angle to the right inside the triangle can be 60
Other angle inside the triangle can be 40
The upper angle outside the triangle 80
The last angle 100
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The figure below shows a shaded circular region inside a larger circle:
Fittoniya [83]

Answer:

Step-by-step explanation:

here you go it in there

6 0
3 years ago
2y • 3 when y = 8 this is so hard
emmainna [20.7K]
2y * 3
2(8) * 3
16 * 3
48
The answer is 48 when y = 8
3 0
3 years ago
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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
How do you solve 2h + 7 > 21
Mrrafil [7]
2h + 7 > 21\\
2h>14\\
h>7
8 0
3 years ago
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Please Help Me On These Math Questions!!!!! Solve Each System By Elimination
nekit [7.7K]
11) -x + y = -1 ; 2x - y = 0
y = -1 + x ; 2x - (-1+x) = 0 ⇒ 2x + 1 - x = 0 ⇒x = -1 
y = -1 + (-1) ⇒ y = -2

12) -2x + y = -20 ; 2x + y = 48
y = -20 + 2x ; 2x + (-20 + 2x) = 48 ⇒ 2x -20 + 2x = 48 ⇒ 4x = 48 + 20
                      4x = 68 ⇒ x = 68/4 ⇒ x = 17
y = -20 + 2(17) ⇒ y = -20 + 34 ⇒ y = 14

13) 3x -y = -2 ; -2x + y = 3
y = 3 + 2x ; 3x - (3 + 2x) = -2 ⇒ 3x - 3 - 2x = -2 ⇒ x = -2 + 3 ⇒ x = 1
y = 3 + 2(1) ⇒ y = 3 + 2 ⇒ y = 5

14) x - y = 4 ; x - 2y = 10
x = 4 + y ; (4 + y) - 2y = 10 ⇒ 4 + y - 2y = 10 ⇒ 4 - y = 10 
                  ⇒ -y = 10 - 4 ⇒ -y = 6 ⇒  y = -6
x = 4 + (-6) ⇒ x = 4 - 6 ⇒ x = -2

15) x + 2y = 5 ; 3x + 2y = 17
x = 5 - 2y ; 3(5-2y) + 2y = 17 ⇒ 15 - 6y + 2y = 17 ⇒ -4y = 17 - 15
               ⇒ -4y = 2 ⇒ y = -2/4 ⇒ y = -1/2
x = 5 - 2(-1/2) ⇒ x = 5 + 2/2 ⇒ x = 5 + 1 ⇒ x = 6 
5 0
3 years ago
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