Answer: Option 1
Step-by-step explanation:
See attached image
Answer:
q ≥ 3
Step-by-step explanation:
add 1 to both sides and there you have your answer!
2 ≤ q-1
+1 +1
3 ≤ q
or
q ≥ 3
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
Answer:
5n^2 - 3n - 8
Step-by-step explanation:
4n.5n - 4n.8 + 4.5n - 4.8
= 20n^2 - 32n + 20n - 32
= 20n^2 - 12n - 32
= 4 ( 5n^2 - 3n - 8 )
= 5n^2 - 3n - 8 [ answer ]
Answer:

Step-by-step explanation:
Let c represents child bikes and a represents adult bikes.
Given : Each child bike requires 4 hours to build and 4 hours to test. Each adult bike requires 6 hours to build and 4 hours to test.
With the number of workers, the company is able to have up to 120 hours of building time and 100 hours of testing time for a week.
Then, the required system of inequality :-

If company make 10 child bikes and 12 adult bikes in the week.
Then Put c=10 and a=12 bikes in (1) and (2).
⇒Bike order meets the restrictions
⇒Bike order meets the restrictions
Hence, the system of inequality best explains whether the company can build 10 child bikes and 12 adult bikes in the week.
