What are the solutions to the equation

Mathematics

1 answer:

sineoko [7]2 years ago

7 0

Answer:

3i , -3i

Step-by-step explanation:

This is correct, I don't see what's wrong.

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A movie theatre sold 1,000 tickets and made $7160 in one night. If an adult tickets costs 9$ and a children's costs 5$, how many

victus00 [196]

540 adult tickets were sold and 460 kid tickets were sold

4 0

2 years ago

SOMEONE HELP ASAPPPP PLEASEEE,PLEASE EXPLAIN HOW U GOT YOUR ANSWER, I NEED AN EXPLANATION IN ORDER TO COMPLETE THIS! NO LINKS OR

slega [8]

Answer:

We must use the formula b x h/2 12 x 8/2 = 48

A=48

We must use the formula 1/2a root c squared - a squared

Solving and substituing will get you 35.78

We must divide 81 by 2 to get 9. Since this is a square, all sides will be 9. Then, we must add 9 four times to get 36 cm as our perimeter

B) If we draw the square with a diagonal line, we can understand that the diagonal line (hypotenus) is s root 2.

3) The formula for this area of a triangle is h x b/2. We must substitute the numbers to get our answer:

h x b /2 = 10 x 20/2 = 200/2 = 100

AREA IS 100cm squared

Step-by-step explanation:

5 0

2 years ago

If c is the curve given by \mathbf{r} \left( t \right = \left( 1 5 \sin t \right \mathbf{i} \left( 1 3 \sin^{2} t \right \mathbf

jonny [76]

With the curve $C$ parameterized by

$C:\mathbf r(t)=\underbrace{15\sin t}_{x(t)}\,\mathbf i+\underbrace{13\sin^2t}_{y(t)}\,\mathbf j+\underbrace{12\sin^3t}_{z(t)}\,\mathbf k$

with $0\le t\le\dfrac\pi2$ , and given the vector field

$\mathbf f(x,y,z)=x\,\mathbf i+y\,\mathbf j+z\,\mathbf k$

the work done by $\mathbf f$ on a particle moving on along $C$ is given by the line integral

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int\limits_{t=0}^{t=\pi/2}\mathbf f(x(t),y(t),z(t))\cdot\frac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$

where

$\mathrm d\mathbf r=(15\cos t\,\mathbf i+26\sin t\cos t\,\mathbf j+36\sin^2t\cos t\,\mathbf k)\,\mathrm dt$

The integral is then

$\displaystyle\int_0^{\pi/2}(15\sin t\,\mathbf i+13\sin^2t\,\mathbf j+12\sin^3t\,\mathbf k)\cdot(15\cos t\,\mathbf i+13\sin2t\,\mathbf j+18\sin t\sin2t\,\mathbf k)\,\mathrm dt$
$=\displaystyle\int_0^{\pi/2}(432\sin^5t\cos t+338\sin^3t\cos t+225\sin t\cos t$
$=269$