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Pepsi [2]
3 years ago
8

If a flask containing one liter of water was poured into 1,000 small cubes, then each cube would contain:

Mathematics
2 answers:
mr Goodwill [35]3 years ago
7 0

Answer: Hello!

You have 1 liter of water separated in 1000 cubes, and you want to know how much contains each cube:

1 liter is equivalent to 1000 mL

then if you divide 1 Liter of water into 1000 equal parts, it has sense to separate it on 1000 cubes of 1 mL each.

But also you can see that 1mL = 1 cm³

Then is equivalent to separate 1 Liter of water in 1000 mL or un 1000  1 cm³.

Then the correct options are A and B.

Gekata [30.6K]3 years ago
3 0
A. 1ml

1L = 1,000ml and 1ml = 1cm³

1,000ml divided by 1,000 = 1ml
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-9/15y-5/15y+3/21=5/15y-5/15y-14/21

-14/15y+3/21=-14/21

Move 3/21 to the other side. Sign changes from +3/21 to -3/21.

-14/15y+3/21-3/21=-14/21-3/21

-14/15y=-14/21-3/21

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-14/15y=-17/21

Multiply both sides by -15/14

-14/15y(-15/14)

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Cross out 14 and 14, divide by 14 then becomes 1

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. The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to nd
Blababa [14]

The question is:

The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution y2(x) of the homogeneous equation

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The second solution y2 is

A(x^4)lnx

Step-by-step explanation:

Given the homogeneous differential equation

x²y'' - 7xy' + 16y = 0

And a solution: y1 = x^4

We need to find a second solution y2 using the method of reduction of order.

Let y2 = uy1

=> y2 = ux^4

Since y2 is also a solution to the differential equation, it also satisfies it.

Differentiate y2 twice in succession with respect to x, to obtain y2' and y2'' and substitute the resulting values into the original differential equation.

y2' = u'. x^4 + u. 4x³

y2'' = u''. x^4 + u'. 4x³ + u'. 4x³ + u. 12x²

= u''. x^4 + u'. 8x³ + u. 12x²

Now, using these values in the original equation,

x²(u''. x^4 + u'. 8x³ + u. 12x²) - 7x(u'. x^4 + u. 4x³)+ 16(ux^4) = 0

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xw' = -w

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w = Ae^(-lnx) (where A = e^C)

w = A/x

But w = u'

So,

u' = A/x

Integrating this

u = Alnx

Since

y2 = uy1

We have

y2 = (Alnx)x^4 = (Ax^4)lnx

Therefore, the second solution y2 is

A(x^4)lnx

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