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timofeeve [1]
2 years ago
6

The water level in the pond decreased from 46 feet to 20 feet. What is the percent decrease in the water level rounded to the ne

arest tenth of the percent?
Mathematics
1 answer:
JulsSmile [24]2 years ago
3 0

We have our equation set up for us:

46-46*x=20

move 20 to the other side an then subtract

46*x=26

divide 46 on both sides

x = 26/46

simplify

x=

13/23 which is around

0.56521..

<h2><u><em>Rounded, the percentage decrease is </em></u></h2><h2><u><em></em></u></h2><h2><u><em>56.5 percent</em></u></h2>

<u><em></em></u>

-Hunter

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Yvonne ran up 1/4 of the total flights of staris in an office building and walked up the rest of she
Korolek [52]
Yvonne ran up 12 flights of stairs.

Explanation:
Understand the situation:

Yvonne ran up 1/4 of the total stairs. Afterwards, she walked the REST of the stairs, which is said to be 9 STAIRS. HOW MANY flights of stairs IN TOTAL did she run up?

When she ran the REST of the stairs, it meant she ran the other 3/4 of stairs because:
1 - 1/4 = 3/4

To add someting, that means that the 9 stairs is 3/4 of the total stairs, so lets fins the total stairs using this information:
9 ÷ 3/4 = 12

LETS CHECK IF 9 IS REALLY 3/4 of 12:
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Graph: y - 10 = -2(z – 10)
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points: (2, 26); (3, 24)

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What is the value of a? A) 28 B) 45 C) 62 D) 73
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Find the differential coefficient of <br><img src="https://tex.z-dn.net/?f=e%5E%7B2x%7D%281%2BLnx%29" id="TexFormula1" title="e^
Gemiola [76]

Answer:

\rm \displaystyle y' =   2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x}

Step-by-step explanation:

we would like to figure out the differential coefficient of e^{2x}(1+\ln(x))

remember that,

the differential coefficient of a function y is what is now called its derivative y', therefore let,

\displaystyle y =  {e}^{2x}  \cdot (1 +   \ln(x) )

to do so distribute:

\displaystyle y =  {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x}

take derivative in both sides which yields:

\displaystyle y' =  \frac{d}{dx} ( {e}^{2x}  +   \ln(x)  \cdot  {e}^{2x} )

by sum derivation rule we acquire:

\rm \displaystyle y' =  \frac{d}{dx}  {e}^{2x}  +  \frac{d}{dx}   \ln(x)  \cdot  {e}^{2x}

Part-A: differentiating $e^{2x}$

\displaystyle \frac{d}{dx}  {e}^{2x}

the rule of composite function derivation is given by:

\rm\displaystyle  \frac{d}{dx} f(g(x)) =  \frac{d}{dg} f(g(x)) \times  \frac{d}{dx} g(x)

so let g(x) [2x] be u and transform it:

\displaystyle \frac{d}{du}  {e}^{u}  \cdot \frac{d}{dx} 2x

differentiate:

\displaystyle   {e}^{u}  \cdot 2

substitute back:

\displaystyle    \boxed{2{e}^{2x}  }

Part-B: differentiating ln(x)•e^2x

Product rule of differentiating is given by:

\displaystyle  \frac{d}{dx} f(x) \cdot g(x) = f'(x)g(x) + f(x)g'(x)

let

  • f(x) \implies   \ln(x)
  • g(x) \implies    {e}^{2x}

substitute

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =  \frac{d}{dx}( \ln(x) ) {e}^{2x}  +  \ln(x) \frac{d}{dx}  {e}^{2x}

differentiate:

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Final part:

substitute what we got:

\rm \displaystyle y' =   \boxed{2 {e}^{2x}   +    \frac{1}{x}  {e}^{2x}  + 2 \ln(x) {e}^{2x} }

and we're done!

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Answer:

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Step-by-step explanation:

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