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andrey2020 [161]
3 years ago
5

Please help with Calc AB

Mathematics
1 answer:
Arturiano [62]3 years ago
6 0

Answer:

Step-by-step explanation:

36.

p(t)=(2t-3)e^{2-t}

p(1)=(2*1-3)e^{2-1}=-e

p(3)=(2*3-3)e^{2-3}=3e^{-1}

average~velocity=\frac{3e^{-1}-(-e) }{3-1} \\=\frac{3+e^{2} }{2e} \\\approx 1.9109\\\approx 1.91

37.

p=(2t-3)e^{2-t} \\v=\frac{dp}{dt}\\=(2t-3)e^{2-t}(-1) +2e^{2-t} \\=(5-2t)e^{2-t}

d(u v)=u dv+v du

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In one year the perseid metor shower had a metor appear every 1/5 5 minutes on average That same year the Leonid metor shower ha
olga55 [171]

In one year, the Perseid meteor shower had a meteor appear every 1 1/5 minutes on average. That same year, the Leonid meteor shower had a meteor appear every 4 2/3 minutes on average. How many more meteors fell during the Perseid meteor shower?

Answer:

325,452 meteors

Step-by-step explanation:

We all know that :

We have 365 days in a year and in each day there are 24 hours, Likewise an hour has 60 minutes

SO, the total number of minutes in a year is :

365 × 24 × 60 = 525600 minutes

For Perseid  meteor:

1 + \frac{1}{5}  \\ \\ = 1 + 0.2 \\  \\ = 1.2

So one meteor fall at 1.2 minutes interval Thus, in a year, we will have :

1 meteor - 1.2 minutes.

x meteor - 525600 minutes.

1.2x = 525600

x = \frac{525600}{1.2}

x = 438000

∴  438,000 meteors fell during the Perseid shower.

For  Leonid meteor shower:

4 + \frac{2}{3} = 4.67

So one meteor fall at 4.67 minutes interval Thus, in a year, we will have :

1 meteor - 4.67 minutes.

x meteors - 525600 minutes.

4.67x = 525600

x = \frac{525600}{4.67}

x = 112548

112,548 meteors fell during the Leonid Shower.

Finally , the numbers of more meteors that  fell during the Perseid meteor shower is calculated by the difference in the number of Perseid meteor shower and Lenoid meteor shower. i.e

438,000 - 112,548 = 325,452

325,452 more meteors fell during the Perseid meteor shower

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%203%20000%20000%5E%7B2%7D%20%20" id="TexFormula1" title=" 3 000 000^{2} " alt=" 3 000 000^{2
Pachacha [2.7K]
{3000000}^{2} + {4000000}^{2} \\ \\ {10}^{6} ( {3}^{2} + {4}^{2} ) \\ \\ {10}^{6} \times 25 \\ \\ 25000000 \: \: \: \: \: Ans.
8 0
3 years ago
How is 400, 000 + 8000 + 500 + 6 written in standard form ?
andreyandreev [35.5K]

Answer:

408,506

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
Three more than two times in a expression
svetlana [45]
3+2x
i think that would be the answer smol child
8 0
2 years ago
Read 2 more answers
Which is the best approximation to a solution of the equation e^x = 2x + 3?
TiliK225 [7]
The correct question is 
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?

we have that

e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term 
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1

then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10

the answer is x=1.10



3 0
3 years ago
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