Answer:
4x+4y=$40 4 pins and 4 rings
Step-by-step explanation:
Making a table is the easiest for this equation.
6/4=10
12/8=20
18/12=30
24/16=40
The '4' marked in the cut-out must be its width, because it can't be the height ... the height of the cut-out must be 3. / / / The area of the entire big rectangle is 12 x 14 = 168. The area of the little cut-out is 4 x 3 = 12. The area left is 168 - 12 = 154 ./ / / By the way ... if you're not getting many answers, it's because your drawings are unreadable. Put away the big orange marker and use a ballpoint pen instead.
What a mysterious expression you have there!
You just have to look at the form of the equations
The only linear equations would be y = mx + b and Ax + By = C
It looks like the ODE is

with the initial condition of
.
Rewrite the right side in terms of the unit step function,

In this case, we have

The Laplace transform of the step function is easy to compute:

So, taking the Laplace transform of both sides of the ODE, we get

Solve for
:

We can split the first term into partial fractions:

If
, then
.
If
, then
.


Take the inverse transform of both sides, recalling that

where
is the Laplace transform of the function
. We have


We then end up with
