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natima [27]
3 years ago
5

What is the solution to the inequality 2(x+9/4)>3

Mathematics
1 answer:
Butoxors [25]3 years ago
4 0

Answer:

-3/4 is the answer

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Find the lowest common denominator of and . A. x3y4 B. x2y3 C. xy4 D. x4y5
Blizzard [7]
Lowest Common Denominator refers to lowes t common multiple. These expressions have two terms 'x' and 'y' and we want to choose the expression that has the highest power such that the other expressions can be multiplied into the common denominator.

For the 'x' term, the highest power is x⁴ and for the 'y' term, the highest power is y⁵

Common denominator of A, B, C, and D: x⁴y⁵




5 0
3 years ago
Select the correct answer. I'll mark brainiest, thank you!
Virty [35]

Answer:

C is your answer.

Step-by-step explanation:

Hope this helps.

8 0
3 years ago
William got scores of q sub 1, , and q sub 3 on three quizzes.
USPshnik [31]

Answer:

a. The average is  =(++) .

b. He can use the formula  =−− q sub 3 , equals 3 x minus , q sub 1 , minus , q sub 2. He will need a score of  =()−−= q sub 3 , equals 3 open 90 close minus 85 minus 88 equals 97 on his third quiz.

Step-by-step explanation:

3 0
3 years ago
How do I subtract 6.45 from 81.3
Aleks [24]
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5 0
3 years ago
A small combination lock on a suitcase has 4 ​wheels, each labeled with the 10 digits 0 to 9. how many 4 digit combinations are
tamaranim1 [39]
This is the easiest way to solve this problem:

Imagine this represents how many combinations you can have for each of the 4 wheels (each blank spot for one wheel): __ __ __ __

For the first situation it says how many combos can we make if no digits are repeated.
We have 10 digits to use for the first wheel so put a 10 in the first slot 
10 __ __ __
Since no digit can be repeated we only have 9 options for the second slot
10 9_ __ __
Same for the third slot, so only 8 options
<u>10</u> <u> 9 </u> <u> 8 </u> __
4th can't be repeated so only 7 options left
<u>10</u> <u> 9 </u> <u> 8 </u> <u> 7 
</u><u>
</u>Multiply the four numbers together: 10*9*8*7 = 5040 combinations


For the next two do the same process as the one above.

If digits can be repeated? You have ten options for every wheel so it would look like this: <u>10</u> <u>10</u> <u>10</u> <u>10
</u>
10*10*10*10 = 10,000 combinations

If successive digits bust be different?
We have 10 for the first wheel, but second wheel only has 9 options because 2nd number can't be same as first. The third and fourth wheels also has 9 options for the same reason.

<u>10</u> <u> 9</u><u> </u> <u> 9 </u> <u> 9 </u>

10*9*9*9 = 7290 combinations






8 0
3 years ago
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