No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.
Consider polynomial equations x^2 = 3 and x^2 = 1.
Obviously they have real solutions.
Subtract the two polynomial equations:
(x^2 - x^2) = (3 - 1)
0 = 2...
We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.
Obviously 0 = 2 has no real solution.
Answer:
180 cars
Step-by-step explanation:
Like you can see in the picture, I like to set up a proportion, so 81 over X and 45 over 100. Then, you can cross multiply, so 81 x 100 = 8100 and then divide, 8100 ÷ 45 = 180, your answer!
Answer:
this is not a real math equation.
Step-by-step explanation:
Answer:
StartFraction negative 1 Over k cubed EndFraction
Step-by-step explanation:
3k / (k + 1) × (k²- 1) / 3k³
= 3k(k² - 1) / (k + 1)(3k³)
= 3k³ - 3k / 3k⁴ + 3k³
= -3k / 3k⁴
= -1/k³
StartFraction k + 1 Over k squared EndFraction
(k + 1) / k²
StartFraction k minus 1 Over k squared EndFraction
(k - 1)/k²
StartFraction negative 1 Over k cubed EndFraction
= -1/k³
StartFraction 1 Over k EndFraction
= 1/k
Hello from MrBillDoesMath!
Answer:
Increasing on the interval [-5, -2]
Discussion:
The function value remains unchanged on [-2, 1] and decreases on [1,8]
Thank you,
MrB