A 15 kg bag of strawberry for $8 ____

1.) What is the equation of the path of firework #1? Write your equation in general form.

valina [46]

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in $y=ax^2+bx+c$

it's easier to find vertex form first then expand to get general form

vertex form is $y=a(x-h)^2+k$ where the vertex is (h,k) and a is a constant

firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

$h_1=a(t-10)^2+50$

to find the value of $a$ , subsitute another point

(0,0)

$0=a(0-10)^2+50)$

$0=100a+50$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_1=\frac{-1}{2}(t-10)^2+50$

expand to get general form

$h_1=\frac{-1}{2}(t^2-20t+100)+50$

$h_1=\frac{-1}{2}t^2+10t-50+50$

$h_1=\frac{-1}{2}t^2+10t$

2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is $h_2=a(t-10)^2+72$

find $a$

use another point

(0,22)

$22=a(0-10)^2+72$

$22=100a+72$

$-50=100a$

$a=\frac{-1}{2}$

so the equation in vertex form is $h_2=\frac{-1}{2}(t-10)^2+72$

3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50

domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

3 0

2 years ago

Given log4(3) =0.7925 and log4(5) =1.1610, use log properties to find log4(300).

Ede4ka [16]

Answer:

4.1145

Step-by-step explanation:

You can use the following formula to solve this kind of problems,

$log_{a}(b) + log_{a}(c) = log_{a}(bc)$

So, let's start of with log4_(300). Since the question have already gave log4_(3) =0.7925 and log4_(5) =1.1610. It means that you have to split the 300 into the simplest form where almost all of the numbers are 3 and 5.

$log_{4}(300) = log_{4}(4 \times 5 \times 3 \times 5) \\ = log_{4}(4) + log_{4}(5) + log_{4}(3) + log_{4}(5) \\ = 1 + 1.1610 + 0.7925 + 1.1610 \\ = 4.1145$