What is equivalent to (1/2)^-2t

Mathematics

2 answers:

oksano4ka [1.4K]3 years ago

7 0

Answer:

4^t

Step-by-step explanation:

Given expression is $(\frac{1}{2})^{-2t}$

To simplify this we apply exponential property

If we have negative exponent at the top then we flip the fraction

that is we take reciprocal of the base 1/2

so 1/2 becomes 2/1 = 2

The expression becomes $2^{2t}$

Now we simplify it further

2^2= 2*2= 4

$2^{2t}=4^t$

vodomira [7]3 years ago

4 0

$( \frac{1}{2} )^{-2t}=2^{2t}=4^t$

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Let's take a look at the first few numbers in the sequence based on the given rule:

$a_{1}=6\\ a_{2}= \frac{1}{4}(6)\\ a_3= \frac{1}{4}\big( \frac{1}{4}\big)6= \big(\frac{1}{4}\big)^2(6)\\ a_{4}= \frac{1}{4} \big(\frac{1}{4}\big)^26= \big(\frac{1}{4}\big)^3(6)$

Inspecting this pattern it seems like the power $\frac{1}{4}$ is being raised to is always one less than the number of the sequence, so if we were on the nth number in the sequence, that part of the expression would be $\big(\frac{1}{4} \big)^{n-1}$ . We also know that we'll be multiplying whatever we get from that by 6, so we can write the full explicit rule for our sequence as

$a_n=6\big( \frac{1}{4} \big)^{n-1}$

Where $a_n$ is the nth number in our sequence.