Answer:
Step-by-step explanation:
We first have to write the equation for the sequence, then finding the first five terms will be easy. Follow the formatting:
and we are given enough info to fill in:
and
and
or in linear format:
where n is the position of the number in the sequence. We already know the first term is -35.
The second term:
so
and
The third term:
and
so
and we could go on like this forever, but the nice thing about this is when we know the difference all we have to do is add it to each number to get to the next number.
That means that the fourth term will be -27 + 4 which is -23.
The fifth term then will be -23 + 4 which is -19. You can check yourself by filling in a 5 for n in the equation and solving:
and
so
This is just simple. For example you have a plane of the form x=a, then you just substitute x with a, and you'll get an equation with y and z only, hence you have a 2-d trace of the intersection. It is just similar for y=b and z=c.
(1) At z=1.5, 2x^2 + 5y^2 + 1.5^2 = 4
2x^2 + 5y^2 = 1.75
Now you have an ellipse in the z=1.5 plane as your trace.
(2) At x=1, 2(1)^2 + 5y^2 + z^2 = 4
5y^2 + z^2 = 2
Now you have an ellipse in the x=1 plane as your trace.
(3) At z=0, 2x^2 + 5y^2 + (0)^2 = 4
2x^2 + 5y^2 = 4
Now you have an ellipse in the z=0 plane as your trace.
(4) At y=0, 2x^2 + 5(0)^2 + z^2 = 4
2x^2 + z^2 = 4
Now you have an ellipse in the y=0 plane as your trace.
Minimal grade to barely pass would be a C, but you should review and go for at least a B or B+ if you want to pass with a C in the class as a whole.
