Speed=distance/time
suppose the distance of the first day is d, and the time is t
distance of the second day: d+0.17d=1.17d
time of the second day: t+0.2t=1.2t
speed of the second day: 1.17d/1.2t=0.975(d/t)=(1-0.025)(d/t)
so the speed of the second day is 2.5% slower than the first day.
Answer:
78.125
Step-by-step explanation:
sorry i can show the work :(
Hi!
So, it looks like you're doing ratios.
Let's look at 9). If the model car is 4 inches long, and 1 inch equals 4 feet, than the actual car is 16 feet long. We know this because 4x4=16.
11) is also just as simple. If 1 cm is 11 km, and the two cities are 99 km apart, then on the map they are 9 cm apart. 11x9 is 99.
13) This is a 1:2 ratio. Basically, the flagpole's shadow is half of its height. You half the man's height to get a 3 foot shadow.
And, finally, 15)- if the model is 4 inches to the real life train's 3 feet, the actual train is 120 ft long, given that the model is 40 in long. 40x3=120.
I hope this helps! :)
In this equation we start by filling in the N with "1" then"2" then "3" and so on
3n+4=
3(1)+4=7
3(2)+4=10
3(3)+4=13
3(4)+4=16
