Answer:By definition, perpendicular line are two lines that intersect at right angles. In other words, the angle made by two lines should be 90°. Therefore, the use of distance formula does not help because it only tells you if the sides are equal. It does not tell you about the intercepted angle.
A technique that can help you to know if two straight lines are perpendicular is is you find their slopes. Let's say the slope of line 1 is m1 and the slope of line 2 is m2. If m1*m2 yields a product of -1, then the lines are perpendicular. This is because if m1 is the negative reciprocal of m2, the lines are perpendicular. But if m1=m2, the lines are parallel, meaning they don't intersect at all.
Therefore, the answer is: Find the slopes and show that their product is -1.
hope it help
Answer:
A function to represent the height of the ball in terms of its distance from the player's hands is 
Step-by-step explanation:
General equation of parabola in vertex form 
y represents the height
x represents horizontal distance
(h,k) is the coordinates of vertex of parabola
We are given that The ball travels to a maximum height of 12 feet when it is a horizontal distance of 18 feet from the player's hands.
So,(h,k)=(18,12)
Substitute the value in equation
---1
The ball leaves the player's hands at a height of 6 feet above the ground and the distance at this time is 0
So, y = 6
So,
6=324a+12
-6=324a


Substitute the value in 1
So,
Hence a function to represent the height of the ball in terms of its distance from the player's hands is 

Here, we want to find the diagonal of the given solid
To do this, we need the appropriate triangle
Firstly, we need the diagonal of the base
To get this, we use Pythagoras' theorem for the base
The other measures are 6 mm and 8 mm
According ro Pythagoras' ; the square of the hypotenuse equals the sum of the squares of the two other sides
Let us have the diagonal as l
Mathematically;
![\begin{gathered} l^2=6^2+8^2 \\ l^2\text{ = 36 + 64} \\ l^2\text{ =100} \\ l\text{ = }\sqrt[]{100} \\ l\text{ = 10 mm} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20l%5E2%3D6%5E2%2B8%5E2%20%5C%5C%20l%5E2%5Ctext%7B%20%3D%2036%20%2B%2064%7D%20%5C%5C%20l%5E2%5Ctext%7B%20%3D100%7D%20%5C%5C%20l%5Ctext%7B%20%3D%20%7D%5Csqrt%5B%5D%7B100%7D%20%5C%5C%20l%5Ctext%7B%20%3D%2010%20mm%7D%20%5Cend%7Bgathered%7D)
Now, to get the diagonal, we use the triangle with height 5 mm and the base being the hypotenuse we calculated above
Thus, we calculate this using the Pytthagoras' theorem as follows;
(x + 3)^3 = (x + 3)(x + 3)(x + 3) = (x + 3)(x^2 + 6x + 9) = x^3 + 9x^2 + 27x + 27
Multiply .6 x 132 and that gets you 79.2 then you add 132 +79.2 =$211.20