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Leni [432]
3 years ago
11

If z is a positive integer, does 4+3(2z-5) represent a number that is greater than, less than, or equal to 2(3z-4)?

Mathematics
1 answer:
Viefleur [7K]3 years ago
5 0

Let us plug in 1 for z. Now we can solve both expressions. 4+3(2x1-5) and 2(3x1-4) and we get that the first expression is equal to -5, and the second expression is equal to -2. So the expression 4+3(2z-5) represents a number that is less than 2(3z-4).

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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

Integrating f_x we get,

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)

Then,

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

Again differentiate with respect to z we get,

f_z=xy+h'(z)=xy+6z

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

3 0
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Answer:

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Step-by-step explanation:

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Sum * Difference = (-29/24)*(-49/24) = 1421/576

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Answer:

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Step-by-step explanation:

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(sorry if it's a bad explanation)

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