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S_A_V [24]
3 years ago
12

Thea has a key on her calculator marked $\textcolor{blue}{\bf\circledast}$. If an integer is displayed, pressing the $\textcolor

{blue}{\bf\circledast}$ key chops off the first digit and moves it to the end. For example, if $6138$ is on the screen, then pressing the $\textcolor{blue}{\bf\circledast}$ key changes the display to $1386$. Thea enters a positive integer into her calculator, then squares it, then presses the $\textcolor{blue}{\bf\circledast}$ key, then squares the result, then presses the $\textcolor{blue}{\bf\circledast}$ key again. After all these steps, the calculator displays $243$. What number did Thea originally enter?
Mathematics
1 answer:
ch4aika [34]3 years ago
4 0

Answer:

$9$

Step-by-step explanation:

Given: Thea enters a positive integer into her calculator, then squares it, then presses the $\textcolor{blue}{\bf\circledast}$ key, then squares the result, then presses the $\textcolor{blue}{\bf\circledast}$ key again such that the calculator displays final number as $243$.

To find: number that Thea originally entered

Solution:

The final number is $243$.

As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,

the number before $243$ must be $324$.

As previously the number was squared, so the number before $324$ must be $18$.

As previously the $\textcolor{blue}{\bf\circledast}$ key was pressed,

the number before $18$ must be $81$

As previously the number was squared, so the number before $81$ must be $9$.

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Helen [10]
All 4 sides are parrell
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3 years ago
A cyclist estimates that he will bike 80 miles this week. He actually bikes 75.5 miles. What is the percent error of the cyclist
maw [93]

Answer:

<em>The percent error of the cyclist's estimate is 5.63%</em>

Step-by-step explanation:

<u>Percentages</u>

The cyclist estimates he will bike 80 miles this week, but he really bikes 75.5 miles.

The error of his estimate in miles can be calculated as the difference between his estimate and the real outcome:

Error = 80 miles - 75.5 miles = 4.5 miles

To calculate the error as a percent, we divide that quantity by the original estimate and multiply by 100%:

Error% = 4.5 / 80 * 100 = 5.625%

Rounding to the nearest hundredth:

The percent error of the cyclist's estimate is 5.63%

5 0
3 years ago
If a and b are positive, then a-b is positive identify the conclusion of the statement
Elanso [62]

Answer:

The statement is wrong. a-b can be positive but not always.

Step-by-step explanation:

There can be three possible situations that are:

  1. a>b
  2. a<b
  3. a=b

Now taking first condition, the result will be positive because a smaller integer is subtracted from a greater one. Such as 5-2=3

In second condition, the result will be negative because a greater integer is subtracted from smaller one. Such as 2-5= -3

In the third condition, the result will always  be zero because both the integers will cancel each other. Such as 5-5=0

6 0
3 years ago
Read 2 more answers
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:

\rm\displaystyle  \frac{d}{dx}  \ln(x)  \cdot  {e}^{2x}  =   \boxed{\frac{1}{x} {e}^{2x}  +  2\ln(x)  {e}^{2x} }

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!

6 0
3 years ago
Helppp please reallly fasttt!!!
Bingel [31]

Answer:

Step-by-step explanation:

3^{2}+ 7^{2} = EF^{2}

58=EF^{2}

EF=\sqrt{58}

4 0
3 years ago
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