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Kruka [31]
3 years ago
10

3x–x–4x+4x What is this simplfied to? Please use exponents.

Mathematics
2 answers:
madam [21]3 years ago
7 0

im pretty sure you dont need exponents for this bud

3x-x-4x+4x

2x+0

2x

sp2606 [1]3 years ago
7 0

2x do you want the steps on how to solve it?


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How do I solve 2t<-4 or 7t>49
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2t<-4
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2. Four pieces of candy are drawn at random from a bag containing five orange pieces and seven brown pieces.
alexandr402 [8]

Answer:

a) 12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495 ways

b) 5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10 ways

c) 7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21 ways

Step-by-step explanation:

Combinatory means combination or arrangement of different elements.

If we have n total elements and we want to find in how many ways we can select x we can use this general formula:

nCx= \frac{n!}{x! (n-x)!}

Where n! = n *(n-1)!

a. How many different ways can four pieces be selected from the 12 colored pieces?

For this case we have:

12C4 = \frac{12!}{4! (12-4)!}= \frac{12!}{4! 8!}=\frac{12*11*10*9*8!}{4! 8!}= 495 ways

b. How many different ways can two orange pieces be selected from five orange pieces?

For this case we have:

5C2 = \frac{5!}{2! (5-2)!}= \frac{5!}{2! 3!}=\frac{5*4*3!}{2! 3!}= 10 ways

c. How many different ways can two brown pieces be selected from seven brown pieces?

For this case we have:

7C2 = \frac{7!}{2! (7-2)!}= \frac{7!}{2! 5!}=\frac{7*6*5!}{2! 5!}= 21 ways

6 0
4 years ago
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Taya2010 [7]

Just follow basic distribution and simplifying rules and your answer is x+46=

8 0
3 years ago
Read 2 more answers
A type of fish for your aquarium costs $9 each. You can spend at most $63.
charle [14.2K]

Answer:

9x>63$

Step-by-step explanation:

nine $ aquarium , at most > 63$

8 0
3 years ago
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