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Alik [6]
3 years ago
14

What is the difference: 3 and 2/3 - 2 and 1/3 in it’s simplest form?

Mathematics
1 answer:
Igoryamba3 years ago
5 0
3 2/3- 2 1/3

3-2= 1

2/3-1/3= 1/3

1 and 1/3 are it's the simplest form.

I hope this helps!
`~kaikers


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What is greater 64kg or 64000g
Aleksandr [31]

Answer: neither

Step-by-step explanation: because 64 kg and 640000g are equal masses.

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Given f(x) = x − 7 and g(x) = x2 . Find g(f(−1)).
ivann1987 [24]

Answer:

64

Step-by-step explanation:

g(f(-1)) can be found by first finding f(-1) then taking that result and putting it into the function g.

f(-1) means we are going to take the expression x-7 and evaluate it for x=-1:  

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So f(-1)=-8.

So we have this so far:

g(f(-1))=g(-8)=(-8)^2=64.

g(-8) was found by replacing x in x^2 with -8.

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Find f(-2)for f(x)=5*3^x
damaskus [11]

f ( - 2 ) = 5 × 3^( - 2 )

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What is the sum of StartRoot negative 2 EndRoot and StartRoot negative 18 EndRoot?
Darya [45]

Answer:

4 \sqrt{2} i

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\sqrt{ - 2}  +  \sqrt{ - 18}

We can rewrite this as

\sqrt{ 2 \times  - 1}  +  \sqrt{ 18 \times  - 1}

This becomes;

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What the recursive formula for the sequence
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\bf n^{th}\textit{ term of an arithmetic sequence} \\\\ a_n=a_1+(n-1)d\qquad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ d=\textit{common difference} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ a_n=2-5(n-1)\implies a_n=\stackrel{\stackrel{a_1}{\downarrow }}{2}+(n-1)(\stackrel{\stackrel{d}{\downarrow }}{-5})


so, we know the first term is 2, whilst the common difference is -5, therefore, that means, to get the next term, we subtract 5, or we "add -5" to the current term.

\bf \begin{cases} a_1=2\\ a_n=a_{n-1}-5 \end{cases}


just a quick note on notation:

\bf \stackrel{\stackrel{\textit{current term}}{\downarrow }}{a_n}\qquad \qquad \stackrel{\stackrel{\textit{the term before it}}{\downarrow }}{a_{n-1}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{current term}}{a_5}\qquad \quad \stackrel{\textit{term before it}}{a_{5-1}\implies a_4}~\hspace{5em}\stackrel{\textit{current term}}{a_{12}}\qquad \quad \stackrel{\textit{term before it}}{a_{12-1}\implies a_{11}}

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