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3 years ago

In △ABC AL is an angle bisector (L∈ BC ). Point M∈ AB so that LM=AM=BM. Find the angles in △ABC, if AC=2AL.

Mathematics

1 answer:

Alex787 [66]3 years ago

3 0

Since LM = AM, point M must be on the perpendicular bisector of AL. Since AM = BM, BL must be perpendicular to AL. This makes ∆ALC a right triangle with hypotenuse AC twice the length of side AL. Hence ∠LAC = ∠LAB = 60°, and AL is angle bisector, median, and altitude.

ΔABC is isosceles with ∠A = 120°, and ∠B = ∠C = 30°.

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Delvig [45]

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.

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3 years ago

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Snowcat [4.5K]

Answer:

just follow the rules

Step-by-step explanation:

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Create the inverse of the coefficient matrix out of the matrix equation. ...

Multiply the inverse of the coefficient matrix in the front on both sides of the equation. ...

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago