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natulia [17]
3 years ago
15

a laboratory, you determine that the density of a certain solid is 5.23×10−6kg/mm3. Convert this density into kilograms per cubi

c meter. Notice that the units you are trying to eliminate are now in the denominator. The same principle from the previous parts applies: Pick the conversion factor so that the units cancel. The only change is that now the units you wish to cancel must appear in the numerator of the conversion factor
Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0

Answer:

5.23\times 10^{-6}\ kg/mm^3=5230\ kg/m^3          

Step-by-step explanation:

Given : The density of a certain solid is 5.23\times 10^{-6} kg/mm³.

To find : Convert this density into kilograms per cubic meter ?

Solution :

We know that,

1\ m^3=10^9\ mm^3

or 10^9\ mm^3=1\ m^3

So, 1 mm^3=\frac{1}{10^9}\ m^3

1 mm^3=10^{-9}\ m^3

Converting 5.23\times 10^{-6} kg/mm³ into kg/m³,

5.23\times 10^{-6}\ kg/mm^3=\frac{5.23\times 10^{-6}}{10^{-9}}\ kg/m^3

5.23\times 10^{-6}\ kg/mm^3=5.23\times 10^{-6+9}\ kg/m^3

5.23\times 10^{-6}\ kg/mm^3=5.23\times 10^{3}\ kg/m^3

5.23\times 10^{-6}\ kg/mm^3=5230\ kg/m^3

Therefore, 5.23\times 10^{-6}\ kg/mm^3=5230\ kg/m^3

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2 years ago
What is greater 1/3 or 1/8
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Nataly [62]
<h3>Answers:</h3>

f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}

================================================

Work Shown:

Part 1

f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out \left(\sqrt{x-1}+5\right)^2

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3 years ago
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