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Iteru [2.4K]
3 years ago
5

Help please. ASAPi dont get it​

Mathematics
1 answer:
charle [14.2K]3 years ago
6 0

Answer:

what do you need help still?

Step-by-step explanation:

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The expression √81 means you need to find a number that, when multiplied by itself, equals 81.What does the expression 81^1/2 me
Nady [450]

the answer is 9 u just need to find two number multipy to 81

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4 years ago
How many times larger is 3.19 x 10^8 than 8.23 x 10^5? SHOW YOUR WORK!!!
svet-max [94.6K]

In order to find the number times something is larger than something, you have to use division where you divide the larger number by the smaller number.

So:

\frac{3.19*10^8}{8.23*10^5} =387.6063183

I'm not sure how much you want to round to,  but the answer is 387.6063183 times.

You can make sure by multiplying the smaller number by 387.6063183 and you should get the larger number:

8.25*10^5*387.6063183=3.19*10^8

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4 years ago
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6 0
3 years ago
I am working on Absolute value and i need help here are the questions
spayn [35]
1) 2x = +/- 4 so x = +/- 2

2) x-3 = 7  OR x - 3 = -7
so x = 10 or x = -4

3)2x + 5 = 21 OR 2x + 5 = -21
so x = 8 or x = -13

4) x + 4 = 6 OR x + 4 = -6
so x = 2 or x = -10

5) x + 2 = -(2x+4) OR x + 2 = 2x + 4
x = -2 or x = -6

6) x + 7 =  0
x = -7

7) you forgot the equal sign somewhere in this problem
4 0
3 years ago
Read 2 more answers
F(x) =
sertanlavr [38]

To make things easier, first rewrite

<em>f(x)</em> = <em>x</em> / (<em>x</em> + 7/<em>x</em>) = <em>x</em> ² / (<em>x</em> ² + 7)

i.e. multiply <em>f</em> by <em>x</em>/<em>x</em>, which is valid because <em>x</em> can't be 0 anyway as it's outside the domain of <em>f</em>.

Now, by the quotient and chain rules, we get

<em />

<em>f '(x)</em> = ((<em>x</em> ² + 7) (<em>x</em> ²)' - <em>x</em> ² (<em>x</em> ² + 7)') / (<em>x</em> ² + 7)²

… = (2<em>x</em> (<em>x</em> ² + 7) - <em>x</em> ² (2<em>x</em>)) / (<em>x</em> ² + 7)²

… = (2<em>x</em> ³ + 14<em>x</em> - 2<em>x </em>³) / (<em>x</em> ² + 7)²

… = 14<em>x</em> / (<em>x</em> ² + 7)²

<em>f ''(x)</em> = ((<em>x</em> ² + 7)² (14<em>x</em>)' - (14<em>x</em>) ((<em>x</em> ² + 7)²)') / ((<em>x</em> ² + 7)²)²

… = (14 (<em>x</em> ² + 7)² - (14<em>x</em>) (2 (<em>x</em> ² + 7)² (<em>x</em> ² + 7)')) / (<em>x</em> ² + 7)⁴

… = (14 (<em>x</em> ² + 7)² - (14<em>x</em>) (2 (2<em>x</em>) (<em>x</em> ² + 7)²)) / (<em>x</em> ² + 7)⁴

… = (14 (<em>x</em> ² + 7)² - 4<em>x</em> (14<em>x</em>) (<em>x</em> ² + 7)²) / (<em>x</em> ² + 7)⁴

… = 14 (<em>x</em> ² + 7)² (1 - 4<em>x</em>) / (<em>x</em> ² + 7)⁴

… = 14 (1 - 4<em>x</em>) / (<em>x</em> ² + 7)²

… = (14 - 56<em>x</em>) / (<em>x</em> ² + 7)²

6 0
3 years ago
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