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3 years ago

Write the value of the expression 2^3/2^3

Mathematics

1 answer:

kiruha [24]3 years ago

6 0

Answer:

Step-by-step explanation:

$\frac{2^3}{2^3}\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\=2^{3-3}\\2^{3-3}=1\\2^{3-3}\\\mathrm{Subtract\:the\:numbers:}\:3-3=0\\=2^0\\\mathrm{Apply\:rule}\:a^0=1,\:a\ne \:0\\=1$

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It cost David $16.75 to fill his 5-gallon gas can.

katen-ka-za [31]

Answer: I divided 16.75 by 5

Step-by-step explanation:

For every 1 gallon hes using 3.35

So 22 x 3.35 is 73.70 so hell need 70 cent more

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3 years ago

Alex and Raul have a total of $5.00 between them. Raul has 60 cents more than Alex. How much does each have?

eimsori [14]

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3 years ago

Un adulto cuya masa de 8kg camina con una rapidez de 130 cm/s

Bas_tet [7]

Answer:

if you need distance then

Step-by-step explanation:

speed = distance /time

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3 years ago

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3 years ago

Read 2 more answers

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

3 years ago