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adell [148]
3 years ago
7

Given: △MOP, Perimeter of △MOP=12+4 sqr rt. 3 m∠P=90°, m∠M=60° Find: MP

Mathematics
2 answers:
Margarita [4]3 years ago
8 0

Consider triangle MOP. This triangle is right triangle, because  m∠P=90°.

Since  m∠P=90°, m∠M=60°, then m∠O=180°-90°-60°=30°.

The leg opposite to the angle with measure 30° is half of the hypotenuse. Let the leg MP=x, then the hypotenuse MO=2x. By the Pythagorean theorem,

PO^2+MP^2=MO^2\Rightarrow PO^2=(2x)^2-x^2=4x^2-x^2=3x^2\Rightarrow PO=\sqrt{3}x.

The perimeter of triangle MOP is

P_{MOP}=MO+PO+MP=x+2x+\sqrt{3}x=3x+\sqrt{3}x.

Since P_{MOP}=12+4\sqrt{3}, then x=4.

Answer: MP=4 units.

atroni [7]3 years ago
7 0

Answer: Mp = 4 units

Step-by-step explanation:

Sorry I just know from RSM, I can't give you a detailed explanation. Hope this Helps!

-Vikii

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Choose 5 cards from a full deck of 52 cards with 13values (2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A) and 4 kinds(spade, diamond, h
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Answer:

a) 182 possible ways.

b) 5148 possible ways.

c) 1378 possible ways.

d) 2899 possible ways.

Step-by-step explanation:

The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question, we have that:

There are 52 total cards, of which:

13 are spades.

13 are diamonds.

13 are hearts.

13 are clubs.

(a)Two-pairs: Two pairs plus another card of a different value, for example:

2 pairs of 2 from sets os 13.

1 other card, from a set of 26(whichever two cards were not chosen above). So

T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182

So 182 possible ways.

(b)Flush: five cards of the same suit but different values, for example:

4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So

T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148

So 5148 possible ways.

(c)Full house: A three of a kind and a pair, for example:

4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).

3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So

T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378

So 1378 possible ways.

(d)Four of a kind: Four cards of the same value, for example:

4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).

1 from the remaining 39(do not involve the kind chosen above). So

T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899

So 2899 possible ways.

4 0
3 years ago
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