Ask question

Ask question

All categories

3 years ago

5

A square piece of paper 160 mm on a side is folded in half along a diagonal. The result is a 45° – 45° – 90° triangle. What is t

he length of the hypotenuse?

1 answer:

posledela3 years ago

6 0

A^2+b^2=c^2

160^2+160^2 = 51,200

Square root of that is <span>226.27416998</span>

You might be interested in

In 2008 Yolanda paid 4873 in social security tax. If the Social Security tax rate was 6.2% to the maximum income and 94,200 that

harina [27]

Answer:

IM A PLAYA FORGIVE ME LOLOL

8 0

2 years ago

Read 2 more answers

What is the product of the 6th square number and the 5th square number?

lukranit [14]

Answer:

900.

Step-by-step explanation:

The first 6 square numbers are 1, 4, 9, 16, 25, 36,

The required product is 25 * 36

= 900.

8 0

2 years ago

What is the probability that a red or green marble will be selected from a bag containing 9 red marbles, 6 blue marbles, 7 green

lara31 [8.8K]

Answer: $\bold{\dfrac{16}{33} = 48\%}$

<u>Step-by-step explanation:</u>

Red or Green

$=\dfrac{red}{total}+\dfrac{green}{total}$

$=\dfrac{9}{33}+\dfrac{7}{33}$

$=\dfrac{16}{33}$

≈ 48%

4 0

3 years ago

The slope between (-3,5) and (6, -4)

gregori [183]

Answer:

-1

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Exercise 10.10.2: Distributing a coin collection. About A man is distributing his coin collection with 35 coins to his five gran

11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

Given that:

all coins are same;

The same implies that the number of the non-negative integral solution of the equation:

$x_1+x_2+x_3+x_4+x_5 = 35$

$x_1 > 0 ; \ \ \ x_1 \ \varepsilon \ Z$

Thus, the number of the non-negative integral solution is:

$^{(35+3-1)}C_{5-1} = ^{39}C_4$

(b)

Here all coins are distinct.

So; the number of distribution appears to be an equal number of ways in arranging 35 different objects as well as 5 - 1 - 4 identical objects

i.e.

$= \dfrac{(35+4)!}{4!}$

$= \dfrac{39!}{4!}$

(c)

Here; provided that the coins are the same and each grandchild gets the same.

Then;

$x_1+x_2+x_3+x_4+x_5 = 35$

$x_1 > 0 ; \ \ \ x_1 \ \varepsilon \ Z$

$x_1=x_2=x_3=x_4=x_5$

$5x_1 = 35\\\\ x_1= \dfrac{35}{5} \\ \\ x_1= 7$

Thus, each child will get 7 coins

(d)

Here; we need to divide the 35 coins into 5 groups, this process will be followed by distributing the coin.

The number of ways to group them into 5 groups = $\dfrac{35!}{(7!)^55!}$

Now, distributing them, we have:

$\mathbf{\dfrac{35!}{(7!)^55!} \times 5!= \dfrac{35!}{(7!)^5}}$

3 0

2 years ago