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Artyom0805 [142]
2 years ago
11

true or false:the points (6,13),(21,33),(99,137)all lie on the-same line. the equation of the line is y=4/3x +5

Mathematics
1 answer:
Anit [1.1K]2 years ago
6 0

The answer is true.

Step-by-step explanation:

To find the points all lie on the same line, we need to substitute the points in the equation of the line, to determine if the values on both sides of the equation are equal.

Substituting the point (6,13) in the equation of the line, we get,

\begin{aligned}y &=\frac{4}{3} x+5 \\13 &=\frac{4}{3}(6)+5 \\&=4(2)+5 \\&=8+5 \\13 &=13\end{aligned}

Thus, the values on both sides are equal. The point (6,13) lie on the same line.

Substituting the point (21,33) in the equation of the line, we get,

\begin{aligned}y &=\frac{4}{3} x+5 \\33 &=\frac{4}{3}(21)+5 \\&=4(7)+5 \\&=28+5 \\33 &=33\end{aligned}

Thus, the values on both sides are equal. The point (21,33) lie on the same line.

Substituting the point (99,137) in the equation of the line, we get,

\begin{aligned}y &=\frac{4}{3} x+5 \\137 &=\frac{4}{3}(99)+5 \\&=4(33)+5 \\&=132+5 \\137 &=137\end{aligned}

Thus, the values on both sides are equal. The point (99,137) lie on the same line.

Thus, all the three points lie on the same plane.

Hence, the answer is true.

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Answer:

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Step-by-step explanation:

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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.

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Answer:

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Step-by-step explanation:

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