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Nimfa-mama [501]
3 years ago
15

What is the answer to -17x=-204

Mathematics
1 answer:
Elenna [48]3 years ago
7 0

The answer for this problem is x=12.

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(10-i)-(12+6i)<br> .............
sergeinik [125]

Answer:

-2 - 7 i

Srry If Im wrng but Hope this helped :)

5 0
3 years ago
Read 2 more answers
The following table shows the assets and liabilities of the Smith family in 2005 and 2009.
Marrrta [24]
2005: 
Asset = 200,000 + 25,000 = 225,000
Liability = 30,000 + 8,000 = 38,000

2009
Asset = 180,000 + 18,000 + 20,000 = 218,000
Liability  = 18,000 + 5,000 = 23,000

Assets decreased by 7,000. From 225,000 to 218,000
Liabilities decreased by 13,000. From 38,000 to 23,000

<span>a. From 2005 to 2009, both assets and liabilities decreased. </span>
6 0
3 years ago
Read 2 more answers
G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =
Svetach [21]
Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j
\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u

So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
=-2\pi a^2(\cos\pi-\cos 0)
=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
3 years ago
How much does he weigh?
Marat540 [252]
He weighs about 200 lb.
8 0
3 years ago
If 5x:(x-2) = 11:3 , find the value of x ?​
Masteriza [31]

Answer:

x = -11/2

Step-by-step explanation:

The ratios need to stay the same

5x        11

----- = -------

x-2       3

Using cross products

5x*3 = (x-2) 11

15x = 11x - 22

Subtract 11 from each side

15x-11x = -22

4x = -22

x = -22/4

x = -11/2

5 0
2 years ago
Read 2 more answers
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