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Maurinko [17]
3 years ago
14

How many ways can Rudy choose 4 pizza toppings from a menu of 14 toppings if each topping can only be chosen once

Mathematics
1 answer:
madam [21]3 years ago
6 0

Answer:

24,024

Step-by-step explanation:

Calculation for How many ways can Rudy choose 4 pizza toppings from a menu of 14 toppings

The 4 pizza toppings from a menu of 14 toppings will be 14,13,12,11

Hence,

14*13*12*11=24,024

Therefore the numbers of way Rudy can choose 4 pizza toppings from a menu of 14 is 24,024

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I think it’s 60 but I could be wrong, sorry.

Step-by-step explanation:

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larisa [96]

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I need help on question 2 i need the simplified answer and the restrictions
Anastaziya [24]

Answer:

x cannot be 5 or -3.

\frac{x+5}{x+3}

Step-by-step explanation:

The restrictions for a fraction is that the bottom cannot be 0.

So if we find when the bottom is 0 we have found the values that x cannot be.

Let's solve x^2-2x-15=0.

Since the coefficient of x^2 is 1 all we have to do is find two numbers whose product is -15 and whose sum is -2.

Those numbers are -5 and 3 since (-5)(3)=-15 and (-5)+(3)=-2.

So the factored form of the equation is:

(x-5)(x+3)=0

This means either x-5=0 or x+3=0.

We do have to solve both.

x-5=0 can be solved by adding 5 on both sides.

x-5+5=0+5

x+0=0+5

x=5

x+3=0 can be solved by subtracting 3 on both sides.

x+3-3=0-3

x+0=0-3

x=-3

So x can be any number except x=5 or x=-3.

We already factored the bottom as (x-5)(x+3).

The top is a difference of squares, x^2-a^2,

which can be factored as (x-a)(x+a).

So the top factors as (x-5)(x+5).

The fraction can then be written as:

\frac{x^2-25}{x^2-2x-15}=\frac{(x-5)(x+5)}{(x-5)(x+3)}

This can further simplified assuming x is not 5 we can write it as \frac{x+5}{x+3}.  I canceled the common factor of (x-5).

5 0
3 years ago
A sequence can be generated by using an=4a(n-1)
WITCHER [35]
This problem is half understanding the question and half just plugging in numbers.

Understanding the Question:
So its saying that n = whole number bigger than one, making n = 2, 3, 4, 5, etc.

Now try plugging any number n into the a_n = 4a_{n-1} equation. Let's use 2 to make it simple. If you plug in 2, you get:
a_2 = 4a_{2-1}\\
a_2 = 4a_1
Remember that the subscripts (the small numbers in the corner) are simply the term number. a_1 is term 1, a_2 is term 2, etc. The equation is saying that "term 2 = 4 times term 1."

If you keep plugging in random numbers for n, you'll see that the equation basically says, "a term = 4 times the term right before it." That makes it easy to plug and chug to get our answer.

Plugging and Chugging
Now we understand the question is asking for the first 4 terms, and each term is 4 times the last term. You're told that term 1, a_1 = 6. That makes the next four terms:

Term 2
a_2 = 4a_{2-1}\\
 a_2 = 4a_1\\
 a_2 = 4(6)\\
 a_2 = 24

Term 3
a_3 = 4a_{3-1}\\ a_3 = 4a_2\\ a_3 = 4(24)\\ a_3 = 96

Term 4
a_4 = 4a_{4-1}\\ a_4 = 4a_3\\ a_4 = 4(96)\\ a_4 = 384

Your first 4 terms are 4, 24, 96, and 384.

--------

Answer: A) F

6 0
3 years ago
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