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Agata [3.3K]
3 years ago
8

I need help with converting liters, millilters, kiloliters etc? Okay, i am really confused! I don't understand how to get 75.01

liters into milliters. I don't understand how to get 367,000 milliliters into kiloliters. I really need help understanding the process of how to do it!? Thanks if you can help me. :)

Mathematics
1 answer:
Mamont248 [21]3 years ago
6 0
Firstly, understand prefixes :)

giga 1000000000
mega 1000000
kilo 1000
deci 0.1
centi 0.01
milli 0.001
micro 0.000001
nano 0.000000001

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The staff commute to work in different ways 2/5 drive, 4/15 catch the bus and 1/3 cycle
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Answer:

2/5, 1/3 4/15

Step-by-step explanation:

3x2/5=6/15

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4/15

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2n-n-4+7n in simplest form
MrMuchimi
<span>2n-n-4+7n in simplest form is found by combining like terms:

(2-1+7)n - 4 = 8n-4, or 4(2n-1)</span>
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3 years ago
At a local store, kelly and martin bought some notebooks and pencils. Kelly bought 4 notebooks and 25 pencils for 16.71. Martin
Nat2105 [25]
Let n = cost of 1 notebook
Let p = cost of 1 pencil

Then,

3n + 4p = 8.5
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You can solve for one variable in terms of the other and then substitute into the remaining equation.

3n + 4p = 8.5
+ 5n + 8p = 14.5

Multiply the top equation by -2 so that the p-containing terms cancel each other out:

-2(3n + 4p = 8.5)
+ 5n + 8p = 14.5
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So after dividing both sides by -1, we see that n = $2.5. Plugging into the first equation gives p = $0.25.


3n + 4p = 8.5
5n + 8p = 14.5
6 0
3 years ago
Consider the following set of vectors. v1 = 0 0 0 1 , v2 = 0 0 3 1 , v3 = 0 4 3 1 , v4 = 8 4 3 1 Let v1, v2, v3, and v4 be (colu
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Answer:

We have the equation

c_1\left[\begin{array}{c}0\\0\\0\\1\end{array}\right] +c_2\left[\begin{array}{c}0\\0\\3\\1\end{array}\right] +c_3\left[\begin{array}{c}0\\4\\3\\1\end{array}\right] +c_4\left[\begin{array}{c}8\\4\\3\\1\end{array}\right] =\left[\begin{array}{c}0\\0\\0\\0\end{array}\right]

Then, the augmented matrix of the system is

\left[\begin{array}{cccc}0&0&0&8\\0&0&4&4\\0&3&3&3\\1&1&1&1\end{array}\right]

We exchange rows 1 and 4 and rows 2 and 3 and obtain the matrix:

\left[\begin{array}{cccc}1&1&1&1\\0&3&3&3\\0&0&4&4\\0&0&0&8\end{array}\right]

This matrix is in echelon form. Then, now we apply backward substitution:

1.

8c_4=0\\c_4=0

2.

4c_3+4c_4=0\\4c_3+4*0=0\\c_3=0

3.

3c_2+3c_3+3c_4=0\\3c_2+3*0+3*0=0\\c_2=0

4.

c_1+c_2+c_3+c_4=0\\c_1+0+0+0=0\\c_1=0

Then the system has unique solution that is (c_1,c_2c_3,c_4)=(0,0,0,0) and this imply that the vectors v_1,v_2,v_3,v_4 are linear independent.

4 0
3 years ago
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