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Arte-miy333 [17]
3 years ago
13

Write the equation of the circle in standard form: Center (7,0) R=1

Mathematics
1 answer:
Mice21 [21]3 years ago
8 0

Answer:

(x-7)^2+y^2=1

Step-by-step explanation:

Hi there!

Equation of a circle: (x-h)^2+(y-k)^2=r^2 where the circle is centered at (h,k) and the radius is r

<u>1) Plug in the given center (7,0)</u>

(x-7)^2+(y-0)^2=r^2\\(x-7)^2+y^2=r^2

<u>2) Plug in the radius (1)</u>

(x-7)^2+y^2=1^2\\(x-7)^2+y^2=1

I hope this helps!

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What could be the intersection of plane PRS and plane QRS?
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The plane PRS passes through the points P, R and S. So it contains the line RS. Also the plane QRS passes through the points Q, R and S. So it contains the line RS as well. Since both the planes contain the line RS, the line RS must be the intersection of plane PRS and QRS

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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
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
A given line has the equation 10x+2y=-2.
Karolina [17]

Answer:

5x+y=12

Step-by-step explanation:

10x+2y=-2=5x+y=12

4 0
3 years ago
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