__7_|_35+63_
_____|__9+5__
35+63=7(9+5)=7<span>14=98 or 35+63=98</span>
We will be using this equation for this problem
d = ut + ½.at²
<span>Given:</span>
<span>initial velocity, u = 0 (falling from rest) </span>
<span>acceleration, a = +9.80 m/s²(taking down as the convenient positive direction) </span>
<span>Time = 1.0s, 2.0s, 3.0s, 4.0s, 5.0s </span>
<span>Using .. d = ½.at² each time (each calculation is the distance from the top) </span>
<span>For 1.0s .. d = ½ x 9.80 x (1²) = 4.90 m </span>
<span>For 2.0s .. d = ½ x 9.80 x (2²) = 19.60 m </span>
<span>3.0s .. d = 44.10m (you show the working for the rest) </span>
<span>4.0s .. d = 78.40 m </span>
<span>5.0s .. d = 122.50m </span>
<span>Plot distance (displacement from the top) on the y-axis against time on the x-axis (label axes and give units for each).The line of best fit will be a smoothly upward curving line getting progressively steeper. Do not join graph points with straight lines.</span>
Answer:
100
Step-by-step explanation:
The function takes in an X value and produces a Y value.
The Y value equals 24 times the X value plus 4 more.
This means that:
Y = 24X + 4
When the X value equals 4, the Y value will be:
Y = 24(4) + 4
Y = 96 + 4
Y = 100
When the X value is 4, the Y value is 100.
Answer:-1
Step-by-step explanation:x>2x+1=x-2x>1
-x>1,x<- 1
Answer:
1.5s+2.5p<20
Step-by-step explanation:
Multiply by each kilo and then make sure it adds up to less than 20!
