1. 250ml
2.2060ml
3. 360 cents
4. 1125ml
5.750ml
6.10
Sorry gtg now hopefully this helps a bit..
Answer:
Monomial - one term
Binomial - two terms
Trinomial - three terms
Polynomial - any polynomial with more than three terms
Brainliest, please :)
Answer:
For f(x) to be differentiable at 2, k = 5.
Step-by-step explanation:
For f(x) to be differentiable at x = 2, f(x) has to be continuous at 2.
For f(x) to be continuous at 2, the limit of f(2 – h) = f(2) = f(2 + h) as h tends to 0.
Now,
f(2 – h) = 2(2 – h) + 1 = 4 – 2h + 1 = 5 – 2h.
As h tends to 0, lim (5 – 2h) = 5
Also
f(2 + h) = 3(2 + h) – 1 = 6 + 3h – 1 = 5 + 3h
As h tends to 0, lim (5 + 3h) = 5.
So, for f(2) to be continuous k = 5
The change in the water vapors is modeled by the polynomial function c(x). In order to find the x-intercepts of a polynomial we set it equal to zero and solve for the values of x. The resulting values of x are the x-intercepts of the polynomial.
Once we have the x-intercepts we know the points where the graph crosses the x-axes. From the degree of the polynomial we can visualize the end behavior of the graph and using the values of maxima and minima a rough sketch can be plotted.
Let the polynomial function be c(x) = x
² -7x + 10
To find the x-intercepts we set the polynomial equal to zero and solve for x as shown below:
x
² -7x + 10 = 0
Factorizing the middle term, we get:
x
² - 2x - 5x + 10 = 0
x(x - 2) - 5(x - 2) =0
(x - 2)(x - 5)=0
x - 2 = 0 ⇒ x=2
x - 5 = 0 ⇒ x=5
Thus the x-intercept of our polynomial are 2 and 5. Since the polynomial is of degree 2 and has positive leading coefficient, its shape will be a parabola opening in upward direction. The graph will have a minimum point but no maximum if the domain is not specified. The minimum points occurs at the midpoint of the two x-intercepts. So the minimum point will occur at x=3.5. Using x=3.5 the value of the minimum point can be found. Using all this data a rough sketch of the polynomial can be constructed. The figure attached below shows the graph of our polynomial.