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Ray Of Light [21]
3 years ago
6

11852.014 rounded to nearest tenth

Mathematics
1 answer:
hammer [34]3 years ago
4 0
It would be 11852 because the 0 cannot be rounded since the one isn't greater than 5.
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Evaluate the expression<br><br>4(5+6)-15<br>​
Alona [7]
Answer 29

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11•4=44
44-15=29
5 0
2 years ago
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svet-max [94.6K]
Solve (x+3) / 3x > 2 . [I guess 3x divides (x+3) and not 3], if so, then
(x+3) > 6x → 3 > 6x-x → OR x < 3/5  OR x<0.6



8 0
3 years ago
Which equation represents the line passing through the points (3, 2) and (–9, 6)?
prisoha [69]
(3,2),(-9,6)
slope = (6 - 2) / (-9 - 3) = -4/12 = -1/3

y = mx + b
slope(m) = -1/3
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now we sub into the formula and find b, the y int
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4 0
3 years ago
I need to solve for this 24+12R &lt; 100
Sindrei [870]

Answer:

here's the answer

Step-by-step explanation:

hope this helps!

6 0
2 years ago
How many three digit numbers have digits whose sum is greater than 2?
LenaWriter [7]

Answer:

896

Step-by-step explanation:

Let's talk first about how many 3 digit numbers there are. The first 3 digit number is 100 and the last is 999. So there are 999-100+1 numbers that are 3 digits long. That simplifies to 900.

Now let's find how many of those have a sum for the digits being 1, then 2 ? Then take that sum away from the 900 to see how many 3 digit numbers have the sum of their digits being more than 2.

3 digit numbers with sum of 1:

The first and only number is 100 since 1+0+0=1.

We can't include 010 or 001 because these aren't really three digits long.

3 digit numbers with sum of 2:

The first number is 101 since 1+0+1=2.

The second number is 110 since 1+1+0=2.

The third number is 200 since 2+0+0=2.

That's the last of those. We could only use 0,1, and 2 here.... Anything with a 3 in it would give us something larger than or equal to 3.

So there are 900-1-3 numbers who are 3 digits long and whose sum of digits is greater than 2.

This answer simplifies to 896.

3 0
3 years ago
Read 2 more answers
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