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3 years ago

The table below shows the relationship between the number of calories burned and the minutes of exercise.

Mathematics

1 answer:

sergejj [24]3 years ago

5 0

Answer:

Option B

Step-by-step explanation:

Since, number of calories 'C' burned is proportional to the number of minutes (m) given to the exercise,

C ∝ m

C = km

Where k is the proportionality constant.

From the table attached,

For C = 250 calories and m = 30 minutes,

250 = 30k

k = $\frac{250}{30}$

= $\frac{25}{3}$

Therefore, equation representing this situation will be,

C = $\frac{25}{3}m$

Option B will be the correct option.

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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,

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

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

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,

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

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Step-by-step explanation:

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