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Sunny_sXe [5.5K]
3 years ago
11

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:
mel-nik [20]3 years ago
6 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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The first three terms of an arithmetic sequence are as follows
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Answer:

32 and 39

Step-by-step explanation:

The common difference d for the sequence is 7

1st term of the sequence a1 is 11.

4th (a4) and 5th  (a5) term of the sequence are required

nth term of an arithmetic sequence an =a1+(n-1)d

a4=11+(4-1)7=32

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4 0
3 years ago
FUNDRAISING A school is raising money by selling calendars for $20 each. Mrs. Hawkins promised a party to whichever of her Engli
____ [38]

Rj, this is the solution to question 19:

• 1st period class sold 60 calendars in total

,

• 2nd period class sold 123 calendars in total

We calculate the average per week, this way:

• 1st period class sold 60/4 = 15 calendars per week

,

• 2nd period class sold 123/4 = 30.75 calendars per week

In consequence the multiplication equation that represents this situation is:

15w + 30.75w = 183, where w is the number of weeks Mrs. Hawkins' classes sold calendars.

This is the solution to question 20:

We already know the average number of calendars that 1st and 2nd period classes sold per week. Now, let's calculate the average for 3nd 4th peiod classes, this way:

• 3rd period class sold 89 calendars in total

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Summarizing, we have:

15 + 30.75 + 22.25 + 31.5 = 99.5 calendars was the average number sold by all her classes in a week.

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7 0
1 year ago
The four corners of a rectangle at the points (-3,-4), (5,-10), (-3,-10), and (5,-4) on a coordinate plane. How many square unit
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Check the picture below, you can pretty much count the units off the grid for the length and width.

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I think C... I'm guessing cause my math is wrong
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How do you solve?3(t+4)-2(2t+3)=-4
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3t+12-4t-6=-4
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8 0
3 years ago
Read 2 more answers
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