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LekaFEV [45]
3 years ago
11

What's 2 to the negative 6 power times 2 to the negative 10th power

Mathematics
2 answers:
mafiozo [28]3 years ago
7 0
When multiplying numbers with exponents, add the exponents.

So the answer is 2^-16
tensa zangetsu [6.8K]3 years ago
5 0
It is about 0.00001525878. You do 2 to the negative 6 power then do 2 to the negative 10th power and multiply those answers together to get about 0.00001525878.
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Expand the following 5(2x-4)
Yuliya22 [10]

Answer:

x= 2

Step-by-step explanation:

5*2= 10 then add the x at the end

5*4=20

then you divide both sides by 10

(10x/10)- (20/10)

x= 2

or if your rewriting it then:

(5*2x)- (5*4)=?

Hope this helped :D

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3 years ago
A bag contains 15 counters:7 red 4 blue and 4 green.a counter is taken out of the bag at random.what is the probability of choos
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|\Omega|=15\\|A|=7+4=11\\\\P(A)=\dfrac{11}{15}\approx73.3\%

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Step-by-step explanation:

 

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3 years ago
Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.
Lilit [14]

For starters,

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Consider the nth partial sum, denoted by S_n:

S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n

Multiply both sides by \frac34:

\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}

Subtract S_n from this:

\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)

Solve for S_n:

-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)

S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)

Now as n\to\infty, the exponential term will converge to 0, since r^n\to0 if 0. This leaves us with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}

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3 years ago
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