Glucose 6 phosphate is regenerated at the end of oxidative phase of pentose phosphate pathway- how it happens explain
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d - 6 (or x = d - 6).
6
Circumference = 2(pi)r OR (pi)d Taking (pi)d = 6 d=6/(pi) d=1.91 (3 significant figures)
You can write this as:(d+2)/6
To find the solution, we first need to calculate the difference between 6 and 8, which is -2. Then, we subtract this difference from d. Therefore, the expression "d decreased by the difference of 6 and 8" can be written as d - (-2), which simplifies to d + 2.