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4 years ago

Find the sum of nth term of the series 1+3/2+5/2^2+7/2^3+.......

Mathematics

1 answer:

ioda4 years ago

6 0

Answer:

an = (1 + 2·(n - 1))/2^(n - 1) = 2^(1 - n)·(2·n - 1)

sn = 6 - 2·0.5^n·(2·n + 3)

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Find cos 30 in terms of cos 0

SVETLANKA909090 [29]

Answer:

cos(30) = (sqrt(3)/2) * cos(0)

Step-by-step explanation:

cos(0) = 1

cos(30) = sqrt(3)/2

Thus, we can write that:

cos(30) = (sqrt(3)/2) * cos(0)

7 0

3 years ago

How do i do KCF ? for 6th grade ;~;

Virty [35]

KCF

Keep

Change

Flip

lets say we are dividing 1/2 and 1/4. you would keep 1/2 the same, change the division sign to a multipluction, and flip 1/4 so that it becomes 4/1. you would then just multiply 1/2 and 4/1

4 0

3 years ago

5a+5b=25 and -5a+5b=35

Alex_Xolod [135]

add the two equations together

5a+5b=25

-5a+5b=35

------------------

0 + 10b =60

divide by 10

b=6

5a + 5b = 25

5a +5(6) = 25

5a +30 =25

subtract 30 from each side

5a =-5

divide by 5

a = -1

Answer (-1,6)

or a=-1 b=6

3 0

3 years ago

The sum of two numbers is 29 and their difference is 13.

harina [27]

Answer:

Sum equation: x + y = 29

Difference equation: x - y = 13

7 0

2 years ago

Simplify this equation please.

Sauron [17]

$\dfrac{\csc^2\theta-3\csc\theta+2}{\csc^2\theta-1}$

Identity:

$\sin^2\theta+\cos^2\theta=1\implies1+\cot^2\theta=\csc^2\theta$

So we can rewrite the denominator to get

$\dfrac{\csc^2\theta-3\csc\theta+2}{\cot^2\theta}$

Multiply numerator and denominator by $\sin^2\theta$ . Several terms will cancel since $\sin\theta\csc\theta=1$ . Also, $\cot\theta=\dfrac{\cos\theta}{\sin\theta}$ . We get

$\dfrac{1-3\sin\theta+2\sin^2\theta}{\cos^2\theta}$

Factorize the numerator, and write $\cos$ in terms of $\sin$ in the denominator to factorize it further to get

$\dfrac{(1-\sin\theta)(1-2\sin\theta)}{\cos^2\theta}=\dfrac{(1-\sin\theta)(1-2\sin\theta)}{1-\sin^2\theta}=\dfrac{(1-\sin\theta)(1-2\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}$

The $1-\sin\theta$ factors cancel, leaving you with

$\dfrac{1-2\sin\theta}{1+\sin\theta}$

which you could simplify a bit further by writing

$\dfrac{1+\sin\theta-3\sin\theta}{1+\sin\theta}=1-\dfrac{3\sin\theta}{1+\sin\theta}$

3 0

4 years ago