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2 years ago

9

Solve the system using an equivalent system

1 answer:

DerKrebs [107]2 years ago

8 0

Add y to both sides in 1st equation
-2x=y
sub -2x for y in2nd equation
3x-(-2x)=10
3x+2x=10
5x=10
divide 5
x=2

sub back
-2x=y
-2(2)=y
-4=y

(x,y)
(2,-4)
C is ans

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Can someome please help me solve this my teacher hasn't done a good with teaching us this.

scoundrel [369]

Answer:

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Step-by-step explanation:

Your calculator can do this for you. You may need to set the display to scientific notation, if that's the form of the answer you want.

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(5·10^2) +(3·10^4)

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2 years ago

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Consider F and C below. F(x, y, z) = yz i + xz j + (xy + 6z) k C is the line segment from (2, 0, −2) to (5, 4, 2) (a) Find a fun

WITCHER [35]

Answer:

Required solution (a) $f(x,y,z)=xyz+3z^2+C$ (b) 40.

Step-by-step explanation:

Given,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k$

(a) Let,

$F(x,y,z)=yz \uvec i +xz\uvec j+(xy+6z)\uvex k=f_x \uvec{i} +f_y \uvex{j}+f_z\uvec{k}$

Then,

$f_x=yz,f_y=xz,f_z=xy+6z$

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$f(x,y,z)=xyz+g(y,z)$

Differentiate this with respect to y we get,

$f_y=xz+g'(y,z)$

compairinfg with $f_y=xz$ of the given function we get,

$g'(y,z)=0\implies g(y,z)=0+h(z)\implies g(y,z)=h(z)$

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$f(x,y,z)=xyz+h(z)$

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on compairing we get,

$h'(z)=6z\implies h(z)=3z^2+C$ (By integrating h'(z)) where C is integration constant. Hence,

$f(x,y,z)=xyz+3z^2+C$

(b) Next, to find the itegration,

$\int_C \vec{F}.dr=\int_C \nabla f. d\vec{r}=f(5,4,2)-f(2,0,-2)=(52+C)-(12+C)=40$

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2 years ago