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<em>div</em><em>ide</em><em> </em><em>the</em><em> </em><em>who</em><em>le</em><em> </em><em>equ</em><em>ation</em><em> </em><em>by </em><em>rt</em><em> </em><em>as</em><em> </em><em>sh</em><em>own</em><em> </em><em>bel</em><em>ow</em>
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<em>then</em><em> </em><em>you</em><em> </em><em>cance</em><em>l</em><em> </em><em>ou</em><em>t</em><em>.</em><em> </em><em>leav</em><em>ing</em>
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<em>which</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>sam</em><em>e</em><em> </em><em>as</em>
<em>
</em>
You times the x and y by 3 good luck
Answer:
cjvnvhhvhjgufvjffuhghhhvc
Might have to experiment a bit to choose the right answer.
In A, the first term is 456 and the common difference is 10. Each time we have a new term, the next one is the same except that 10 is added.
Suppose n were 1000. Then we'd have 456 + (1000)(10) = 10456
In B, the first term is 5 and the common ratio is 3. From 5 we get 15 by mult. 5 by 3. Similarly, from 135 we get 405 by mult. 135 by 3. This is a geom. series with first term 5 and common ratio 3. a_n = a_0*(3)^(n-1).
So if n were to reach 1000, the 1000th term would be 5*3^999, which is a very large number, certainly more than the 10456 you'd reach in A, above.
Can you now examine C and D in the same manner, and then choose the greatest final value? Safe to continue using n = 1000.
Answer:
14x+14/33
Step-by-step explanation:
14÷3/11+14x=
