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Natasha_Volkova [10]
3 years ago
5

0.002840909 to the nearest tenth

Mathematics
1 answer:
timurjin [86]3 years ago
3 0
The nearest tenth is still zero. Reason being you have to start from the right, and once you round the far most 9 to the zero it turns to a one which is too small of a number to round up.
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I’m Really lost if I could get a answer it would be greatly appreciated
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

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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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4 0
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Substitute

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