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Gnoma [55]
3 years ago
11

Plz plz help me for math

Mathematics
1 answer:
brilliants [131]3 years ago
4 0

Answer:

Oasis idiot and a good time in the day before

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How do I write a rule for the nth term of the sequence?
crimeas [40]

first of all we have the 5th term and 2nd term

T5=768 =a*r^4

T2=-12=a*r ^1

make a subject of formula in the both equations

that is a=T2/r and T5/r^4

a=12/r and a=768/r^4

when u equate them u will get

-64=r^3 so r=-4 then to find a that is 1st term

since second term is T2=a*r=12 an r=-4 then

a=12/-4 hence a=-3

4 0
3 years ago
Here is another one...
IgorC [24]
The correct answer is C. :)
4 0
3 years ago
Find the limit............​
liubo4ka [24]

Answer: -\frac{9}{x^2}

====================================================

Work Shown:

\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9}{x+\Delta x} - \frac{9}{x}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x}{x(x+\Delta x)} - \frac{9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{\frac{9x-9(x+\Delta x)}{x(x+\Delta x)}}{\Delta x}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{9x-9x-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\

\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9\Delta x}{x\Delta x(x+\Delta x)}\\\\\\\displaystyle \lim_{\Delta x \to 0^{+}} \frac{-9}{x(x+\Delta x)}\\\\\\\displaystyle \frac{-9}{x(x+0)}\\\\\\\displaystyle -\frac{9}{x^2}\\\\\\

The trick here is to first simplify the expression so that the \Delta x term in the very lower denominator cancels out. This is to avoid dividing by zero. Once this division by zero error is avoided, we can replace the delta x term with 0 and evaluate the limit as \Delta x approaches 0 from the right or positive direction.

5 0
3 years ago
(Please help!!)
vladimir1956 [14]
L=5+2w would be the best answer
7 0
3 years ago
Which number is rational?
irina [24]
D
hope this helps
this is because the .3 is repeating but it is the same number repeating(it is a pattern)
3 0
4 years ago
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