1. You already did it.
2. Table
3. t (years since 1990)
4. n (# of cigarettes sold)
5. (t, n)
6. You can see the distribution of the data pretty neatly. There are also many more advantages including it's easier to calculate standard deviation, easier to see the mean, mode and median, and it's also much easier to just tell the extrema of the dataset by just looking at the scattergram.
# shared / # being shared = 5/6 Each sister gets 5/6 of a pizza.
Answer:
x^2+5x-14
Step-by-step explanation:
you multiply both equations so that you get:
x^2-2x+7x-14
and then you simplify it to:
x^2+5x-14
<em><u>If you found this helpful please give it a thanks </u></em>
<em><u>I would also appreciate it if you give me a brainliest</u></em>
<u>Annotation</u>General formula for distance-time-velocity relationship is as following
d = v × t
The velocity of the first car will be v₁, the time is 2 hours, the distance will be d₁.
The velocity of the second car will be v₂, the time is 2 hours, the distance will be d₂.
One of them traveling 5 miles per hour faster than the others. That means the velocity of the first car is 5 miles per hour more than the velocity of the second car.
v₁ = v₂ + 5 (first equation)
The distance of the two cars after two hours will be 262 miles apart. Because they go to opposite direction, we could write it as below.
d₁ + d₂ = 262 (second equation)
Plug the d-v-t relationship to the second equationd₁ + d₂ = 262
v₁ × t + v₂ × t = 262
v₁ × 2 + v₂ × 2 = 262
2v₁ + 2v₂ = 262
Plug the v₁ as (v₂+5) from the first equation2v₁ + 2v₂ = 262
2(v₂ + 5) + 2v₂ = 262
2v₂ + 10 + 2v₂ = 262
4v₂ + 10 = 262
4v₂ = 252
v₂ = 252/4
v₂ = 63
The second car is 63 mph fast.Find the velocity of the first car, use the first equationv₁ = v₂ + 5
v₁ = 63 + 5
v₁ = 68
The first car is 68 mph fast.
Answer


Answer:
hfjf fnnc v jud dbhj ejem emd jfbf
Step-by-step explanation: