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3 years ago

11

What is the square root of -1

1 answer:

MAXImum [283]3 years ago

5 0

Answer:

i

Step-by-step explanation:

$\sqrt{-1}$ is the same thing as i, which is an imaginary number.

Since you can't take the square root of negative numbers, we use i to do it.

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Chords AB and CD intersect at point E, AE = 10, EB = 4, and CE = 8. Therefore, ED =

Genrish500 [490]

So here is how we are going to find out what is ED.
Based on the given figure, it states that, AE is 10, and EB is 4 and CE is 8.
So, <span>(AE/CE)=(ED/EB)
10/8 = ED/4 <<multiply both sides by the common denominator which is 8 and the result would be:
80/8 = 8ED/4
10 = 2ED <<divide both sides by 2 and we get
ED = 5.
Therefore, the measurement of ED is 5.
Hope this answer helps. Let me know if you need more help next time!</span>

5 0

3 years ago

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Find the value of x.<br> helpppp

Bezzdna [24]

Answer: $55^{\circ}$

Step-by-step explanation:

The measure of the third unknown arc is

$360^{\circ}-70^{\circ}-110^{\circ}=180^{\circ}$ .

So, $x=\frac{180^{\circ}-70^{\circ}}{2}=\boxed{55^{\circ}}$

8 0

2 years ago

I need help, please. it would mean a lot I need it now

fomenos

Answer:

60 feet

Step-by-step explanation:

a - 2x

b - x

c - 3x

120/6 = 20

20 x 3 = 60

8 0

4 years ago

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it takes Dan 32 minutes to complete 2 pages . at this rate how many pages does he complete in 200 minutes

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3 years ago

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A particle is moving along the x-axis so that its position at t ≥ 0 is given by s(t)=(t)ln(5t). Find the acceleration of the par

lyudmila [28]

Answer:

$a(\frac{1}{5e})=5e$

Step-by-step explanation:

we are given equation for position function as

$s(t)=tln(5t)$

Since, we have to find acceleration

For finding acceleration , we will find second derivative

$s'(t)=\frac{d}{dt}\left(t\ln \left(5t\right)\right)$

$=\frac{d}{dt}\left(t\right)\ln \left(5t\right)+\frac{d}{dt}\left(\ln \left(5t\right)\right)t$

$=1\cdot \ln \left(5t\right)+\frac{1}{t}t$

$s'(t)=\ln \left(5t\right)+1$

now, we can find derivative again

$s''(t)=\frac{d}{dt}\left(\ln \left(5t\right)+1\right)$

$=\frac{d}{dt}\left(\ln \left(5t\right)\right)+\frac{d}{dt}\left(1\right)$

$=\frac{1}{t}+0$

$a(t)=\frac{1}{t}$

Firstly, we will set velocity =0

and then we can solve for t

$v(t)=s'(t)=\ln \left(5t\right)+1=0$

we get

$t=\frac{1}{5e}$

now, we can plug that into acceleration

and we get

$a(\frac{1}{5e})=\frac{1}{\frac{1}{5e}}$

$a(\frac{1}{5e})=5e$

5 0

3 years ago