All categories

3 years ago

PLEASE ANSWER DONT IGNORE ): I WILL GIVE U BRAINLIEST AND THANKS AND 5 STARS

Mathematics

1 answer:

Lera25 [3.4K]3 years ago

7 0

Answer:

Niether of them are correct

Step-by-step explanation:

If you actually put into your calculator 2^28, it equals 268,435,456. This is because you're not just squaring 28, you are multiplying 2 by 2-- 38 times. This will quickly add up, even if you start out with 2 it will create a very large number.

You might be interested in

What are the domain and range of f (x) = log x minus 5? domain: x > 0; range: all real numbers domain: x 5; range: y > 5 d

tresset_1 [31]

Answer:

domain: x>5 Range: y>-5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Select the correct answer.

Inessa [10]

I think its a :) if not c

6 0

2 years ago

Expand the following 5( 2x -4 )

ycow [4]

Answer:

10x-20

Step-by-step explanation:

5(2x-4)

5x2=10

5x4=20

8 0

3 years ago

Meneca bought a package of 435 party favors to give to the guests at her birthday party. She calculated that she could give 9 pa

Makovka662 [10]

She is expecting approximately 48 guests

8 0

3 years ago

Read 2 more answers

5. Find the general solution to y'''-y''+4y'-4y = 0

CaHeK987 [17]

For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

$(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0$

Simplify to,

$e^{yt}(y^3-y^2+4y-4)=0$

Then find solutions,

$\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}$

For non repeated real root y, we have a form of,

$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

and,

$y_3=a-bi$

the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

Hope this helps.

r3t40

5 0

4 years ago