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2 years ago

What does this multiple by 38 , 51 , 64 , 77 , 90

Mathematics

1 answer:

nadezda [96]2 years ago

3 0

Answer:

It's an arithmetic sequence. Instead of multiplying, you're adding 13. The answer is Tn=13n+25

Step-by-step explanation:

If you minus 38 from 51 (51-38), you get 13 and this 13 is constant throughout.

It's the same if you do (64-51).

Therefore 13 is our difference, we call it d.

Next you use the formula Tn=a+(n-1)d and substitute.

a is the first term of our sequence therefore it's a=38 and d=13.

We don't know n which is the number of terms.

When we sub its gonna be Tn=38+(n-1)(13)

And then you solve.

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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,

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

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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,

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

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,

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

(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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