This describes a parabolic path that starts at x=0 on a value of y = -5.1. That's when the rock is below ground in the hole. To find the answer we need to know the second x axis crossing. The first is when it is rising and reaches ground level, the second is after it peaks and falls back to the ground and lands. Using the quadratic formula I get the two zeros of this function to be about
x= 15.29, and 66.71. The rock then lands 66.71 feet from him horizontally.
Answer:
A
Step-by-step explanation:
Given
y = x² + 2
Substitute the x value from an ordered pair from each table into the equation and compare the value obtained with the y value of the ordered pair
Table A
using (- 2, 6 )
y = (- 2)² + 2 = 4 + 2 = 6 ← this corresponds to the y value of the ordered pair
Table B
Using (- 2, - 2)
y = (- 2)² + 2 = 4 + 2 = 6 ≠ - 2 ← thus not a solution
Table C
Using (- 2, - 2)
y = (- 2)² + 2 = 4 + 2 = 6 ≠ - 2 ← thus not a solution
Thus Table A gives the solutions to y = x² + 2
Divide 45 by 5 and 5 hours by 5 so it’s 9 miles per hour. So the equation would be: m = 9/h (a fraction)
Answer:
A. 871.97 m
Step-by-step explanation:
Model this is a <u>right triangle</u>.
Use the <u>sine trigonometric ratio</u> to find the height of the weather balloon from the ground.
<u>Sine trigonometric ratio</u>
where:
is the angle- O is the side opposite the angle
- H is the hypotenuse (the side opposite the right angle)
Given:
- = 33°
- O = height
- H = 1601 m
Substitute the given values into the formula and solve for O:
