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3 years ago

1010101201010001 in binary

Mathematics

1 answer:

lesantik [10]3 years ago

6 0

Answer:

Its not possible theres no 2´s in binary

Step-by-step explanation:

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Can you prove it??<br>it's hard,I tried but couldn't solve this.

BARSIC [14]

I'll abbreviate $s=\sin\theta$ and $c=\cos\theta$ , so the identity to prove is

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)$

On the left side, we can simplify a bit:

$\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}$

$\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}$

Then

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)$

So the establish the original equality, we need to show that

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s$

Combine the fractions:

$\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}$

We can rewrite the denominator as

$c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1$

then using the fact that $c^2+s^2=\cos^2\theta+\sin^2\theta=1$ , we get

$1-2s^2+2s-1=2s-2s^2$

so that we have

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s$

as desired.

6 0

4 years ago

PLEASE HELP FAST

Igoryamba

5 + (-10) = 5 - 10
= -5 .

5 + (-3) = 5 -3
=2 .

5 + 3 = 8 .

5 + 0 = 5 .

8 0

4 years ago

Help me prove that 2 + 2 is fish.. HELP PLS

Artist 52 [7]

Answer:

NO 2+2 = 22

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

It costs $45 for 9 baseballs. How much would it cost to buy two baseballs?

Step2247 [10]

Answer: It would cost $10 for 2 baseballs

Step-by-step explanation:

1.) You take 45 and divide it by 9 to find the cost of 1 baseball

2.) You then take the cost of a baseball and multiply it by 2

3 0

3 years ago

The range of f(x)=5x2^x is all positive real numbers true or false

creativ13 [48]

Answer: true
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6 0

4 years ago