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Marrrta [24]
2 years ago
14

What is the unit digit of 7^7⁷?​

Mathematics
1 answer:
pishuonlain [190]2 years ago
8 0

Answer:

Hello,

Step-by-step explanation:

unit mean the unit digit of ...

unit(7^0)=unit(1)=1\\unit(7^1)=unit(7)=7\\unit(7^2)=unit(49)=9\\unit(7^3)=unit(7^2)*7=unit(9*7)=unit(63)=3\\unit(7^4)=unit(7^3)*7=unit(3*7)=1\ just\ like\ unit(7^0)\\

\left \{ \begin {array}{cc}1&if\ n \in 4*\mathbb{N}\\7&if\ n \in 4*\mathbb{N}+1\\9&if\ n \in 4*\mathbb{N}+2\\3&if\ n \in 4*\mathbb{N}+3\\\end {array} \right.\\\\\\unit(7^{7^{7}})=unit(7^{(7^{7})})=unit(7^{823543})=unit(7^3)=3\\\\unit((7^{7})^{7}) =unit(7^{49} )=unit(7^{1})=7\\

I have given 2 answers since i don't know what you mean with 7^7⁷.

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Use any method to solve the equation. If necessary, round to the nearest hundredth. 9x^2-21=0
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Answer:

x = 1.53, -1.53.

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Help me out with this
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Let's follow the transformations that happen to A, to get to A' and A''.

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It moves to (-5, 2) which is where A' is located. Note the x coordinate stays the same while the y coordinate flips from negative to positive. This must mean we applied a reflection over the x axis.

That rule in general is (x, y) \rightarrow (x,-y)

------------------------------------

Now compare A'(-5,2) and A''(1,4). We can shift A' 6 units to the right and then 2 units up so we move from A' to A''.

Algebraically this is stated as (x,y) \rightarrow (x+6, y+2)

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Applying that rule to B'(-1,2) gets us to

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