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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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Step-by-step explanation: The rational expression in this case is undefined if the the denominator is equal to zero, which means that the x cannot be whatever makes the denominator zero, so:

x² - 81 = 0 ⇒ x² = 81 ⇒ x = +/- √81 ⇒ x = +/- 9

Note that we are using the equal sign just because this particular problem askes for a solutions that is undefined.

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Step-by-step explanation:

To express this as a ratio, we simply place 15 over 20

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15/20

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3 years ago
A sled is being held at rest on a slope that makes an angle theta with the horizontal. After the sled is released, it slides a d
Alenkasestr [34]

Answer:

μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

d₂ = (d₁*Sin θ) / μ

Step-by-step explanation:

a) We apply The work-energy theorem

W = ΔE

W = - Ff*d

Ff = μ*N = μ*m*g

<em>Distance 1:</em>

- Ff*d₁ = Ef - Ei

⇒  - (μ*m*g*Cos θ)*d₁ = (Kf+Uf) - (Ki+Ui) = (Kf+0) - (0+Ui) = Kf - Ui

Kf = 0.5*m*vf² = 0.5*m*v²

Ui = m*g*h = m*g*d₁*Sin θ

then

- (μ*m*g*Cos θ)*d₁ = 0.5*m*v² - m*g*d₁*Sin θ  

⇒   - μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ   <em>(I)</em>

 

<em>Distance 2:</em>

<em />

- Ff*d₂ = Ef - Ei

⇒  - (μ*m*g)*d₂ = (0+0) - (Ki+0) = - Ki

Ki = 0.5*m*vi² = 0.5*m*v²

then

- (μ*m*g)*d₂ = - 0.5*m*v²

⇒   μ*g*d₂ = 0.5*v²     <em>(II)</em>

<em />

<em>If we apply (I) + (II)</em>

- μ*g*Cos θ*d₁ = 0.5*v² - g*d₁*Sin θ

μ*g*d₂ = 0.5*v²

 ⇒ μ*g (d₂ - Cos θ*d₁) = v² - g*d₁*Sin θ   <em>  (III)</em>

Applying the equation (for the distance 1) we get v:

vf² = vi² + 2*a*d = 0² + 2*(g*Sin θ)*d₁   ⇒   vf² = 2*g*Sin θ*d₁ = v²

then (from the equation <em>III</em>) we get

μ*g (d₂ - Cos θ*d₁) = 2*g*Sin θ*d₁ - g*d₁*Sin θ

⇒  μ (d₂ - Cos θ*d₁) = Sin θ * d₁

⇒   μ =  Sin θ * d₁ / (d₂ - Cos θ*d₁)

b)

If μ is a known value

d₂ = ?

We apply The work-energy theorem again

W = ΔK   ⇒   - Ff*d₂ = Kf - Ki

Ff = μ*m*g

Kf = 0

Ki = 0.5*m*v² = 0.5*m*2*g*Sin θ*d₁ = m*g*Sin θ*d₁

Finally

- μ*m*g*d₂ = 0 - m*g*Sin θ*d₁   ⇒   d₂ = Sin θ*d₁ / μ

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