PLEASE HELP FAST!!! AND EXPLAIN YOUR WORK

Write an exponential decay function to model the situation. Compare the average rates of change over the given intervals.

Firlakuza [10]

The exponential decay function could be the model of the situation comparing what I said do not utilize because the user is trying to acoses the points

6 0

3 years ago

PLEASE HELP !! 40 POINTS ! ILL GIVE BRAINLIEST

xenn [34]

The answer is NOQ and MLJ

3 0

3 years ago

Someone please.. help me in this please I don’t want to fail it..please

igomit [66]

Answer:

So it wants you to first find the mean (average) of the data set, then find the distance everything is from the mean (average).

Step-by-step explanation:

3+5+9+14+16+18= 65

65 divided by the amount of numbers you put in (6) = 10.83 rounded to the nearest hundreth

So now we know the mean (average) now we need to see how far away each number is from the mean (average).10.83-3 is 7.83. 10.83-5 is 5.83. 10.83-9 is 1.83. 10.83-14 is -14.17, but -14.17 would be a normal 14.17 because it wants the absolute value. 10.3-16 is -5.17 (of course keep it as a positive), and finally 10.83-18 is -7.17 as a positive. Now for the mean absolute value of the data they likely mean from 0, so it would be 10.83.

Hope it helped and wasn't too confusing :)

7 0

3 years ago

martin mixes 1 cup lemonade with 4 cups cranberry juice to make his favorite drink how much cranberry juice does he need if he u

mars1129 [50]

He will need 20 cups of cranberry juice.

3 0

3 years ago

Read 2 more answers

I pull the throttle in my racing plane at a = 12.0 m/s2. i was originally flying at v = 100. m/s. where am i when t = 2.0, t = 5

Misha Larkins [42]

The distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.

We have,

Acceleration of plane i.e. a = 12 m/s²

And,

Velocity of plane i.e. v = 100 m/s

And,

t₁ = 2s

t₂ = 5s

t₃ = 10s

So,

Now,

Using the equation of motion,

i.e.

S = vt - $\frac{1}{2}$ at²

Here,

S = Distance,

v = initial velocity,

a = acceleration

t = time taken

Now,

For t₁ = 2s,

Putting values in above equation we get,

S = (100 * 2) - ( $\frac{1}{2}$ * 12 * 2²)

On solving we get,

S = 200 - 24 = 176 m,

So,

Plane will be at 176m distance.

Now,

For t₂ = 5s,

Putting values in above equation we get,

S = (100 * 5) - ( $\frac{1}{2}$ * 12 * 5²)

On solving we get,

S = 500 - 150 = 350 m,

So,

Plane will be at 350m distance.

Now,

For t₃ = 10s,

Putting values in above equation we get,

S = (100 * 10) - ( $\frac{1}{2}$ * 12 * 10²)

On solving we get,

S = 1000 - 600 = 400 m,

So,

Plane will be at 400m distance.

Hence, we can say that the distance covered by plane when t = 2s will be 176m and when t = 5s will be 350m and when t = 10s will be 400m found using equation of motion.

To learn more about equation of motion click here

brainly.com/question/12114762

#SPJ4

4 0

1 year ago