Please, What do you think of this. Please I need help on this

Mathematics

1 answer:

stich3 [128]3 years ago

8 0

Answer:

Step-by-step explanation:

Since Year 3 and 4 have the highest students just subtract 33-25 and you get 8

You might be interested in

Determine whether or not the vector field is conservative. if it is, find a function f such that f = ∇f. if the vector field is

amm1812

$\dfrac{\partial f}{\partial x}=3y^2z^3\implies f(x,y,z)=3xy^2z^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=6xyz^3+\dfrac{\partial g}{\partial y}=6xyz^3$
$\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)$

$\dfrac{\partial f}{\partial z}=9xy^2z^2+\dfrac{\mathrm dh}{\mathrm dz}=9xy^2z^2$
$\implies\dfrac{\mathrm dh}{\mathrm dz}=0\implies h(z)=C$

$\implies f(x,y,z)=3xy^2z^3+C$

The potential function exists, so $\mathbf f$ is indeed conservative.

7 0

4 years ago

Find the area between y = 8 sin ( x ) y=8sin⁡(x) and y = 8 cos ( x ) y=8cos⁡(x) over the interval [ 0 , π ] . [0,π]. (Use decima

Marina86 [1]

Answer:

0.416 au

Step-by-step explanation:

Let y1=8sin(x) and y2=8cos(x), we must find the area between y1 and y2

$\int\limits^\pi _0{(8cos(x)-8sin(x))} \, dx = 8\int\limits^\pi _0{(cos(x)-sin(x))} \, dx =\\8(sin(x)+cos(x)) evaluated(0-\pi )=\\8(sin(\pi )-sin(0))+8(cos(\pi )-cos(0))=\\8(0.054-0)+8(0.998-1)=8(0.054)+8(-0.002)=0.432-0.016=0.416$