All categories

2 years ago

What is the length of MO, given that figure LMNO is a square?

Mathematics

2 answers:

fgiga [73]2 years ago

5 0

Answer:

8 (C)

Step-by-step explanation:

The diagonals of a square are congruent and bisect each other.

One half of a diagonal = 4

Therefore, the whole diagonal is 8.

Minchanka [31]2 years ago

3 0

Answer:

Step-by-step explanation:

putting it simple without any terms, half of the line LN is 4, all the other lines inside the rectangle are equal, so just double the length for the length of MO for the answer, 8

You might be interested in

What is the domain of the function in the graph above?

alexdok [17]

Answer: [-1,infinity)

Step-by-step explanation:

4 0

3 years ago

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.

4 0

2 years ago

RSTU is a rectangle with vertices at R(-3,1), S(1 ,1), T(-3,3) and U (1, 3). Describe two lines of reflection that would carry t

miss Akunina [59]

Answer:

The two lines of reflection are $x=-1$ and $y=2$

Step-by-step explanation:

I will attach a file with the graph of the rectangle and the lines of reflection.

We know that RSTU is a rectangle with vertices at $R=(-3,1)$ , $S=(1,1)$ ,

$T=(-3,3)$ and $U=(1,3)$

The first step is to draw the vertices on the plane and them the rectangle.

The rectangle is a symmetrical figure. Therefore if we want to carry the rectangle onto itself using a line of reflection this line must goes through the centroid of the rectangle.

Given that we have a rectangle whose width is $a$ and its length is $b$ its centroid will be place at $(\frac{a}{2},\frac{b}{2})$ .