There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/- 
x = +/- i
NAP = 180 - NAC - PAT = 180 - 61 - 46 = 119 - 46 = 73 degrees
Answer:
the decimal equivalent is 0.125
Answer:
Y = -1/2x+4
Step-by-step explanation:
The Y intercept is 4, so that's where the +4 comes in.
The slope is rise over run, so you count how many up you go from a whole number, in this case one, and that goes on top. Then you count how many it goes over, in this case two, and that goes on the bottom. The negative comes from which way the slope is going; down or up.
This revolves around exact trig values - no easy way to say this, you just need to memorise them. They are there for sin cos and tan, but I will give you the main tan ones below - note this is RADIANS (always work in them when you can, everything is better):
tan0: 0
tanpi/6: 1/sqrt(3)
tanpi/4: 1
tanpi/3: sqrt(3)
tanpi/2: undefined
Now we just need to equate -2pi/3 to something we understand. 2pi/3 is 1/3 of the way round a circle, so -2pi/3 is 1/3 of the way round the circle going backwards (anticlockwise), so on a diagram we already know it's in the third quadrant of the circle (somewhere between pi and 3pi/2 rads).
We also know it is pi/3 away from pi, so we are looking at sqrt(3) or -sqrt(3) because of those exact values.
Now we just need to work out if it's positive or negative. You can look up a graph of tan and it'll show that the graph intercepts y at (0,0) and has a period of pi rads. Therefore between pi and 3pi/2 rads, the values of tan are positive. Therefore, this gives us our answer of sqrt(3).
