Answer:

Step-by-step explanation:
In this problem, it is given that,
The stopping distance, D, in feet of a car is directly proportional to the square of it's speed, V.
We need to write the direct variation equation for the scenario above. It can be given by :

To remove the constant of proportionality, we put k.

k is any constant
Hence, this is the required solution.
Answer:
No it is not a solution
Step-by-step explanation:
To solve this, you should replace your (x,y) with (8,3)- 8 being your x value and 3 being your y value.
So this inequality, your y value is supposed to be bigger than the value of whatever 3x-4 equals.
So, your inequality should be 3>3(8)-4
3*8=24-4=20
So is 3>(greater than)20? No. Which then means that (8,3) is not a solution because it would make the inequality not true.
Hope this helps!
8.006
-6.38 = 1.626
Hope this help
Let h(t)= 0 (because when it reaches the ground h=0)
0=-16t^2+116
16t^2=116
divide both sides by 16
t^2= 7.25
square root both sides to find t
t=2.69sec
Answer:
x<±1 and x< -3
Step-by-step explanation:
Given the inequality x^2-1/3x+9<0, we are to find the values of x that satisfies the inequality
x^2-1/3x+9<0
x^2-1/3x+9 (3x+9)²<0* (3x+9)²
(x^2-1)(3x+9) < 0
x²-1 < 0 and 3x + 9 <0
x²-1 < 0
x²<1
x<±√1
x<±1
Also 3x + 9 <0
3x < -9
x < -9/3
x < -3
Hence the required values of x are x<±1 and x< -3