Given A and B are points in nth dimensions.
E.g. (a1, a2, ..., an), (b1, b3, ..., bn)
The vector from point A to point B is given by B-A.
Taking the modulus or magnitude of this will give us the distance.
Therefore distance = || B-A ||
It looks like you have the domain confused for the range! You can think of the domain as the set of all "inputs" for a function (all of the x values which are allowed). In the given function, we have no explicit restrictions on the domain, and no situations like division by 0 or taking the square root of a negative number that would otherwise put limits on it, so our domain would simply be the set of all real numbers, R. Inequality notation doesn't really use ∞, so you could just put an R to represent the set. In set notation, we'd write

and in interval notation,

The <em>range</em>, on the other hand, is the set of all possible <em>outputs</em> of a function - here, it's the set of all values f(x) can be. In the case of quadratic equations (equations with an x² term), there will always be some minimum or maximum value limiting the range. Here, we see on the graph that the maximum value for f(x) is 3. The range of the function then includes all values less than or equal to 3. As in inequality, we can say that
,
in set notation:

(this just means "f(x) is a real number less than or equal to 3")
and in interval notation:
![(-\infty,3]](https://tex.z-dn.net/?f=%20%28-%5Cinfty%2C3%5D%20)
1/4 is the slope!
That equation is in slope intercept form aka y=mx+b. M is the slope, and b is the y intercept
In that equation, m=1/4 therefore 1/4 is the slope
Hope this helps!
Answer: 8ac+12b+4
Step-by-step explanation: Simplify the expression.
Hope this helps you out! ☺
To find the height that the rocket has reached, we can use the following equation to calculate distance.
s = ut + 1/2 at²
s - distance - h
u - initial velocity - 64 ft/s x 0.3048 m/ft = 19.5 m/s
a - acceleration, since the rocket is travelling upwards against gravitational acceleration, its said to be deceleration. therefore value is -9.8 ms⁻²
t - time taken - t seconds
substituting values in the equation
h = 19.5 ms⁻¹ x t s - 1/2 x 9.8 ms⁻² x (t s)²
h = 19.5t - 4.9t²