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3 years ago

How many ways can you distribute 4 different balls among 4 identical boxes?

Mathematics

2 answers:

ludmilkaskok [199]3 years ago

5 0

Answer: 24 different ways

Step-by-step explanation:

Hope this helped!

<!> Brainliest is appreciated! <!>

Aloiza [94]3 years ago

4 0

Answer:

24 different ways

Step-by-step explanation:

You have to make a choice which ball to put in the first box. You have 4 options.

Then you must make a choice which ball to put in the second box. You have 3 options (there are three balls left)

Then you have to make a choice which ball to put in the third box You have 2 options (there are two balls left)

It isn't necessary to make a choice about the fourth box, because there's only 1 option (there's only one ball left)

So you have 4 choices for the first box.

For each of those 4 choices, you have 3 choices for the second box. That makes 12 number of ways.

For each of those 12 ways, you have 2 choices left for the third box. That makes 24 number of ways.

You have only 1 ball left for the fourth box, so that leaves you at ................. ....... 24 number of ways.

Hope this helps

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How do you solve -28-6x=-2(1-3x)-2

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3 years ago

Read 2 more answers

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

Step-by-step explanation:

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

Proof LHS → RHS

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$

$\text{Identity:}\qquad \qquad \dfrac{\sin^2 A}{1-\cos A}$

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4 years ago

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laiz [17]

Answer:

(2,10) or x=2 y=10

Step-by-step explanation:

1. Pick one of your equations and solve for a variable. I chose the first equation and solved for x.

5x-2y=-10 (Move the -2y to the other side, you need to do the opposite so you add +2y to -10)

5x=2y-10 (Divide the 5 from the x)

x=2/5y-2

2. Now take what you got for x and plug it into the x variable on the other equation.

3(2/5y-2)+6y=66 (Multiply 3 by 2/5y and -2)

6/5y-6=6y=66 (Move the -6 to the other side and add 6/5y to 6y)

36/5y=72 (Since the number on the y is a fraction, you must do the opposite to the other side)

y=72/1 x 5/36 (Flip your fraction and multiply it by the 72)

y=10

3. Now that you have one of the variables solved for, in order to get the other we must plug in what we have to the first equation.

5x-2(10)=-10 (Multiple 2 by 10)

5x-20=-10 (Move -20 to the other side, since you do the opposite add +20 to the -10)

5x=10 ( Divide 10 by 5)

x= 2

4. If needed, plug in the values of x and y to check your solution.

Hope this could help! :)

3 0

3 years ago