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Three less than twice the third of three consecutive integers equals the sum of the three integers. Find the integers...... Where do i even start? 1+2+3=(3*2-3)?
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Let the three consecutive integers be: N, N + 1, N + 2 "Three less than twice the third of three consecutive integers equals the sum of the three integers." 2(N+2) - 3 = N + N + 1 + N + 2 solve for N.
i see, thank you very much
You are welcome. What are you getting for N?
i got -2
Yes, the three integers are: -2, -1 and 0.
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hey, how would i simplify \[\sqrt{4-x^2}\]
a^2 - b^2 = (a+b)(a-b) 4 - x^2 = 2^2 - x^2 = (2+x)(2-x) sqrt(4-x^2) = sqrt((2+x)(2-x))
thank you (:
can i jus take the twos out from there?
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