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Mathematics

2 answers:

Tatiana [17]3 years ago

3 0

Answer:

Hiiiiiii

The answer is -36

rodikova [14]3 years ago

3 0

Brainly?
-36

have a good day

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Write the rule for the nth term of the following sequence.

denis-greek [22]

Answer:

$2^{n-3}$ is the nth term of the given sequence.

Step-by-step explanation:

The sequence given is

$\frac{1}{4},\frac{1}{2},1,2,4 .....$

We can clearly see that 1st term is $\frac{1}{4}$ and 2nd term is $\frac{1}{2}$

2nd term is obtained by multiplying the 1st term by 2.

2nd term is $\frac{1}{2}$ and 3rd term is 1.

3rd term is obtained by multiplying the 2nd term by 2.

3rd term is 1 and 4th term is 2.

4th term is obtained by multiplying the 3rd term by 2.

Clearly, the given series is a Geometric Progression(GP) with

First term, $a = \frac{1}{4}$

Common Ratio, $r = 2$

We know that $n^{th}$ term for a GP is:

$a_n = ar^{n-1}$

Putting values of a and r

$a_n = \dfrac{1}{4}2^{n-1}\\a_n = \dfrac{1}{2^{2} }2^{n-1}\\a_n = 2^{n-1-2}\\a_n = 2^{n-3}$

Hence, $2^{n-3}$ is the nth term of the given sequence.

3 0

3 years ago

√(25 - 10 √(3) + 3) (not as a decimal)

nikitadnepr [17]

Answer:

$\huge\boxed{\sqrt{25-10\sqrt3+3}=5-\sqrt3}$

Step-by-step explanation:

$\sqrt{25-10\sqrt3+3}=(*)\\\\25=5^2\\3=(\sqrt3)^2\\10\sqrt3=(2)(5)(\sqrt3)\\\\(*)=\sqrt{5^2-(2)(5)(\sqrt3)+(\sqrt3)^2}\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\=\sqrt{(5-\sqrt3)^2}\\\\\text{use}\ \sqrt{a^2}=|a|\\\\=|5-\sqrt3|=5-\sqrt3\\\\^{\text{because}\ 5>\sqrt3\to5-\sqrt3>0}$