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2 years ago

Determine the relative order of the metric prefixes of kilo-, micro-, centi-, and milli-.

Mathematics

1 answer:

nadya68 [22]2 years ago

8 0

The relative order of the metric prefixes is micro-, milli-, centi- and kilo-,

<h3>What are metric units?</h3>

Metric units are units used to measure quantities

The units are given as:

kilo-, micro-, centi-, and milli-

In metric units, the value of each prefix is:

Kilo = 1000

Micro = 0.0001

Centi = 100

Milli = 0.1

In order of arrangement, we have:

Micro = 0.0001

Milli = 0.1

Centi = 100

Kilo = 1000

Hence, the relative order of the metric prefixes is micro-, milli-, centi- and kilo-,

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Bumek [7]

Answer:

9/10 I do this a lot

step-by-step explanation:

1/2 changes to 5/10 and 2/5 becomes 4/10 then you add and there is your answer

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3 years ago

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3 years ago

For the composite function, identify an inside function and an outside function and write the derivative with respect to x of th

alexira [117]

Answer:

The inner function is $h(x)=4x^2 + 8$ and the outer function is $g(x)=3x^5$ .

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

Step-by-step explanation:

A composite function can be written as $g(h(x))$ , where $h$ and $g$ are basic functions.

For the function $f(x)=3(4x^2+8)^5$ .

The inner function is the part we evaluate first. Frequently, we can identify the correct expression because it will appear within a grouping symbol one or more times in our composed function.

Here, we have $4x^2+8$ inside parentheses. So $h(x)=4x^2 + 8$ is the inner function and the outer function is $g(x)=3x^5$ .

The chain rule says:

$\frac{d}{dx}[f(g(x))]=f'(g(x))g'(x)$

It tells us how to differentiate composite functions.

The function $f(x)=3(4x^2+8)^5$ is the composition, $g(h(x))$ , of

outside function: $g(x)=3x^5$

inside function: $h(x)=4x^2 + 8$

The derivative of this is computed as

$\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=3\frac{d}{dx}\left(\left(4x^2+8\right)^5\right)\\\\\mathrm{Apply\:the\:chain\:rule}:\quad \frac{df\left(u\right)}{dx}=\frac{df}{du}\cdot \frac{du}{dx}\\f=u^5,\:\:u=\left(4x^2+8\right)\\\\3\frac{d}{du}\left(u^5\right)\frac{d}{dx}\left(4x^2+8\right)\\\\3\cdot \:5\left(4x^2+8\right)^4\cdot \:8x\\\\120x\left(4x^2+8\right)^4$

The derivative of the function is $\frac{d}{dx}\left(3\left(4x^2+8\right)^5\right)=120x\left(4x^2+8\right)^4$ .

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3 years ago

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3 years ago

The sum of four consecutive odd integers is three more than five times the least of the integers. Find the integers.

Olegator [25]

Answer:

Step-by-step explanation:

{k - some integer}

2k+1 - the first odd integer (the least)

5(2k+1) - five times the least

5(2k+1)+3 -<u> three more than five times the least</u>

2k+1+2 = 2k+3 - the odd integer consecutive to 2k+1

2k+3+2 = 2k+5 - the next odd consecutive integer (third)

2k+5+2 = 2k+7 - the last odd consecutive integer (fourth)

2k+1+2k+3+2k+5+2k+7 - <u>the sum of four odd consecutive integers</u>

2k+1 + 2k+3 + 2k+5 + 2k+7 = 5(2k+1) + 3

8k + 16 = 10k + 5 + 3

- 10k -10k

-2k + 16 = 8

-16 - 16

-2k = -8

÷(-2) ÷(-2)

k = 4

2k+1 = 2•4+1 = 9

2k+3 = 2•4+3 = 11

2k+5 = 2•4+5 = 13

2k+7 = 2•4+7 = 15

Check: 9+11+13+15 = 48; 48-3 = 45; 45:5 = 9 = 2k+1

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3 years ago