Answer:
The speed of light in a vacuum is 186,282 miles per second
Step-by-step explanation:
The speed of light in a vacuum, commonly denoted c, is a universal physical constant that is important in many areas of physics.Its exact value is defined as 299 792 458 metres per second (approximately 300 000 km/s or 186 000 mi/s). It is exact because, by international agreement, a meter is defined as the length of the path traveled by light …
kilometers per hour: 1080000000
meters per second: 299792458
miles per hour: 671000000
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.
Answer:
≈ 1.32471795725...
Explanation:
If x is one less than its cube, then
x = x³ - 1,
x³ - x - 1 = 0
so f(x) = ? would be an appropriate (continuous) function to apply the Intermedate Value Theorem on some appropriate interval to see if it takes on the value 0 in that interval.
f(x) = x³ - x - 1
For large x the left hand side is positive, for x = 0 it is negative. The root can be calculated exactly, it is given by:
≈ 1.32471795725...
The picture is a bit blurry but what I saw was 8a^3+27=0
8a^3+27=0 first you move 27 to the right side of the equation
8a^3=-27 then you divide both sides by 8
a^3 = -27/8 then take the cubed root of both sides
a= -3/2
Hope that helps ;)
Number one:
A=Pounds of apples
B=Pounds of oranges
0.75a + 0.89b
All I could figure out was number one, so sorry!