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Oksanka [162]
3 years ago
8

1 . a person who weighs 100 pounds on earth weighs 16.6 ib on the moon. part

Mathematics
1 answer:
timurjin [86]3 years ago
3 0
M = E/6

a) 185/6 lb = 30.8333... lb
b) 300 lb
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Jim rented a truck for one day. There was a base fee of $17.99, and there was an additional charge of 76 cents for each mile dri
dalvyx [7]

Answer:

287 miles

Step-by-step explanation:

The base fee was $17.99 and an additional 76 cents ($0.76) for every mile.

We can represent the amount he will pay for rent with the equation below:

C = 17.99 + 0.76m

where m = number of miles driven

Jim paid $236.11, therefore:

236.11 = 17.99 + 0.76m

0.76m = 236.11 - 17.99

0.76m =218.12

m = 218.12 / 0.76 = 287 miles

Jim drove 287 miles.

5 0
3 years ago
14) x - 27 0<br> 3<br> X^3-27=0<br> X^3-27=0
Inessa [10]

Answer:

For the top equation, x = 27 (I think you meant to put x - 27 = 0. If not, I can always edit my answer for you)

For the bottom, x = 3

Hope this helps! Have a grrrreat day lol

5 0
3 years ago
Julie and Jenna are selling bracelets that they made at a craft store. They are hoping to sell 200 bracelets. If Julie can make
raketka [301]
10 since 12+8 is 20 and thats daily 200/20=10
4 0
3 years ago
Read 2 more answers
"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e
Art [367]

Answer:

a. v(t)= -6.78e^{-16.33t} + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)=e^{\int\limits^  {}k \, dt } =e^{kt}. We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ e^{kt}v' + ke^{kt}v = ge^{kt} ⇒ [ve^{kt}]' = ge^{kt}. Integrating, we have

∫ [ve^{kt}]' = ∫ge^{kt}

    ve^{kt} = \frac{g}{k}e^{kt} + c

    v(t)=   \frac{g}{k} + ce^{-kt}.

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + ce^{-16.33 * 0} = 16.33 + c

       c = 9.55 -16.33 = -6.78.

So, v(t)=   16.33 - 6.78e^{-16.33t}. m/s = - 6.78e^{-16.33t} + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78e^{-16.33 x 0.5} + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

6 0
3 years ago
What is this plz help me
iren [92.7K]
20 is the answer for x
8 0
3 years ago
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