1.) what is the equation, in standard form, of the path of firework #1?

2.) what is the equation, in vertex form, of the path of firework #2?

3.) what is the domain and range of firework #1?

Part 1)

Part 2)

Part 3) The domain of firework #1 is the interval (-∞,∞) and the range is the interval (-∞,50]

Step-by-step explanation:

1.) What is the equation, in standard form, of the path of firework #1?

we know that

The equation of a vertical parabola in vertex form is equal to

where

a is a coefficient

(h,k) is the vertex

In this problem we have

The vertex is the point (10,50)

substitute

we have the roots

(0,0) and (20,0)

For x=0,y=0

substitute in the equation and solve for a

substitute

Convert to standard form

A quadratic equation in standard form is equal to

so

2.) What is the equation, in vertex form, of the path of firework #2?

we know that

The equation of a vertical parabola in vertex form is equal to

where

a is a coefficient

(h,k) is the vertex

In this problem we have

The vertex is the point (10,72)

substitute

we have the root

(22,0)

For x=22,y=0

substitute in the equation and solve for a

substitute

3.) What is the domain and range of firework #1?

we know that

The domain of firework #1 is the interval (-∞,∞)

All real numbers

The range of firework #1 is the interval (-∞,50]

All real numbers less than or equal to 50

ddsf

step-by-step explanation:

dfs

10.416 meters per second

step-by-step explanation:

distance=1 00 meters

time= 9.6 seconds in decimal form or 9 3/5 as a mixed number

average speed= 100/9.6

=10.416 meters per second or 10 52/125 as a mixed number.