How do you get to 1,525 when you have 378 521 339

1 answer:

7 0

I don't know if I fully understand the question, but here's the easy way out:
378,521,339-1,525 = 378,519,814, so 378,521,339-378,519,814 = 1,525 is your answer.

You might be interested in

What is the point-slope form of a line with slope 3/2 that contains the point (-1 2)

Drupady [299]

Answer:

The point-slope form of this equation is y - 2 = 3/2(x + 1)

Step-by-step explanation:

To find the answer to any equation in point slope form, start with the base form of the equation.

y - y1 = m(x - x1)

Now input the slope for m and the point at (x1, y1)

y - 2 = 3/2(x + 1)

3 0

3 years ago

A store can buy 8 cartons of milk for 24 dollars or 7 cartons of milk for $28.

larisa86 [58]

Answer:

Step-by-step explanation:

28/7=4

24/8=3

8 0

2 years ago

) Set up a double integral for calculating the flux of F=3xi+yj+zk through the part of the surface z=−5x−2y+2 above the triangle

Fynjy0 [20]

The surface (call it $S$ ) is a triangle with vertices at the points

$x=0,y=0\implies z=2\implies(0,0,2)$

$x=0,y=2\implies z=-2\implies(0,2,-2)$

$x=2,y=0\implies z=-8\implies(2,0,-8)$

Parameterize $S$ by

$\vec s(u,v)=(1-v)(2,0,-8)+v\bigg((1-u)(0,2,-2)+u(0,0,2)\bigg)=(2-2v,2v-2uv,-8+6v+4uv)$

with $0\le u\le1$ and $0\le v\le1$ . Take the normal vector to $S$ to be

$\vec s_v\times\vec s_u=(20v,8v,4v)$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_v\times\vec s_u)\,\mathrm du\,\mathrm dv$