Answer:
0 <=t<=21
Step-by-step explanation:
Projectile is Moving upwards on an interval of (0 to 21), if we plot Velocity vs Time and denote positive y-axis above 0 and negative y-axis below 0(for velocity), then from 0 to 21 t projectile is moving upwards and has positive velocity, when the projectile reaches the top of it's motion and returns back down to ground it's velocity is negative and is plotted below the y =0 (note that is for t > 21).
hence for the interval 0 <=t <=21 the instantaneous velocity is positive (Note, instantaneous velocity is also the derivative of the velocity or the slope ).
The perimeter of a shape is the sum of the lengths of its sides.
So, to find the perimeter of this quadrilateral, all we have to do is add the side lengths and simplify.
(x² - 6) + (2x + 5) + (x² - 3x) + (4x² + 2x)
x² + x² + 4x² + (-3x) + 2x + 2x + (-6) + 5
6x² + (-3x) + 2x + 2x + (-6) + 5
6x² + x + (-6) + 5
6x² + x + (-1)
6x² + x - 1
So, the perimeter of the quadrilateral is the quantity (6x² + x - 1).
Hope this helps!
Answer:
-1/2
Step-by-step explanation:
In a linear relationship, the rate of change of one variable with respect to the other is <em>constant</em>. When we talk about <em>change</em>, we're looking for a <em>difference</em> of values.
If we look at the first and second rows, the change in x is 1 - (-1) = 2, while the change in y is 9 - 10 = -1. Usually we refer to these changes as Δx and Δy (read like "delta-x" and "delta-y"), and the <em>rate of change </em>is the number we get by dividing one of these by the other.
The rate of change we're used to seeing, sometimes called the <em>slope</em>, is Δy/Δx. So, using the values we've already found:

Answer:
see below
Step-by-step explanation:
by rewriting the function we can see that it has a minimum at
![y=7x^2+7x-7\Rightarrow y=7(x^2+x-1)\Rightarrow y=7[(x+\frac{1}{2})^2-\frac{1}{4}-1]\Rightarrow](https://tex.z-dn.net/?f=y%3D7x%5E2%2B7x-7%5CRightarrow%20y%3D7%28x%5E2%2Bx-1%29%5CRightarrow%20y%3D7%5B%28x%2B%5Cfrac%7B1%7D%7B2%7D%29%5E2-%5Cfrac%7B1%7D%7B4%7D-1%5D%5CRightarrow)
![y=7[(x+\frac{1}{2})^2-\frac{5}{4}]\Rightarrow minimum \ (-\frac{1}{2}, -\frac{35}{4})](https://tex.z-dn.net/?f=y%3D7%5B%28x%2B%5Cfrac%7B1%7D%7B2%7D%29%5E2-%5Cfrac%7B5%7D%7B4%7D%5D%5CRightarrow%20minimum%20%5C%20%28-%5Cfrac%7B1%7D%7B2%7D%2C%20-%5Cfrac%7B35%7D%7B4%7D%29)
bye.