Sunday, June 7, 2015
Solve the inequality and express the solution in interval form.
Solution:
For convenience we'll compare an appropriate expression to 0 instead of 1. Then we can find x-values of interest to determine where to shade (i.e. where the inequality is true).
This graph shows our solution another way. Notice that we graph both sides of the inequality. Then we can actually see when the left side is greater...

Find the 11th term of the sequence 1,2,4,8,…
Friday, June 5, 2015
Sunday, May 31, 2015
What is the solution set for the given equation? 3|x + 4| = 18
Let's try a different approach this time...
3|x + 4| = 18
has the same solution as
|x + 4| = 6
We look for the number(s) that can add/subtract to obtain 6. We can see that x=2 will work. No other positive number will work, however if we consider negative numbers then -10 seems possible. In fact is does work because |-10+4| = |-6| = 6. So x=-10.
3|x + 4| = 18
has the same solution as
|x + 4| = 6
We look for the number(s) that can add/subtract to obtain 6. We can see that x=2 will work. No other positive number will work, however if we consider negative numbers then -10 seems possible. In fact is does work because |-10+4| = |-6| = 6. So x=-10.
What is the number of turning points in the graph of the function of x defined below?
y = 2x2 + 5x - 7
Solution:
The polynomial y has degree n=2 so the maximum number of turns it can have is n-1 = 2-1 = 1. (In fact we know that our polynomial is a quadratic and it opens upwards so this is another way to know it only has one turn).
The polynomial equation x(x2+4)(x2-x-6)=0 has how many real roots?
Solution:
x2+4 contributes complex zeros. This means that the factors x and x2-x-6 contribute real zeros. Therefore the polynomial has 3 real zeros
Saturday, May 30, 2015
Friday, May 29, 2015
if tan(theta) = -3 and cos(theta) < 0, then tan(theta/2) = ?
so

so

(we take the negative case because cosine is negative for this example)
now

Therefore...

which simplifies to

so
now
Therefore...
which simplifies to
Subscribe to:
Posts (Atom)


















