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belka [17]
3 years ago
14

Determine the value of x for the following system of equations.

Mathematics
2 answers:
nikklg [1K]3 years ago
4 0
The answer is C) 6.
umka21 [38]3 years ago
4 0
You’re answer would be C, or 6
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As part of a board game, players choose 5 unique symbols from 9 different symbols to create their secret password. How many diff
frozen [14]

Answer:

15,120

Step-by-step explanation:

For the first symbol, there are 9 options to choose from. Then 8, then 7, and so on. Since each player chooses 5 symbols, they will have a total of 9\cdot 8 \cdot 7 \cdot 6\cdot 5=\boxed{15,120} permutations possible. Since the order of which they choose them matters (as a different order would be a completely different password), it's unnecessary to divide by the number of ways you can rearrange 5 distinct symbols. Therefore, the desired answer is 15,120.

8 0
3 years ago
Read 2 more answers
HELP me look at pics
Setler [38]
M+n+3+m+n+4
=3m+2n+7

answer
C. 3m+2n+7
3 0
4 years ago
Read 2 more answers
Dillon pays x dollars to get a haircut. He gives the stylist a
Svetlanka [38]

Answer:

C. 1.2x

Step-by-step explanation:

He pays 1 x for his haircut, and an additional 20% on top of that, or an additional 0.2x. If you write that as an expression, you can simplify it like this:

1x+0.2x\\1.2x

You can also work it out by multiplying by 1.2 as stated in answer C. If his haircut costs $10, adding a 20% tip of $2 is the same as multiplying by 1.2.

10\times1.2=12

3 0
2 years ago
What is 5 more than the quotient of a number and 6 equals 7
loris [4]

Answer:

x/6 + 5 = 7

x/6 = 2

x = 12

Step-by-step explanation:

6 0
3 years ago
PLS HELP! NO GUESSING<br><br> 5r + r3s<br><br> IS IT A MONOMIAL YES OR NO EXPLAIN ANSWER
Natalka [10]
Nope, <span>5r + r3s is not a monomial. 
Monomials cannot include addition or subtraction, and since this uses addition, it is not a monomial. 
</span>

I hope this helps and have a great day! If you need anymore help you can link me to another question and I will try to solve it!

6 0
3 years ago
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