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Burka [1]
3 years ago
14

I need this to be solved plzzzz

Mathematics
1 answer:
tatiyna3 years ago
4 0

Answer:

ln2

Step-by-step explanation:

n --> infinity

Integral of 1/N as N varies from n+1 to 2n

Integral of 1/N = ln(N)

Upper limit - lower limit

ln(2n) - ln(n + 1)

ln[2n/(n + 1)]

ln[2/(1 + 1/n)]

As n --> infinity,

1/n --> 0

ln[2/(1 + 1/n)] --> ln2

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What is the equation of the line through (5, 1) with a slope of -3?
Vlad1618 [11]

Answer:

B. y-1=3(x-5)

Step-by-step explanation:

y=_+_(x-_)

y=1+(-3)(x-5)

y-1=3(x-5)

6 0
3 years ago
a park is 4 times as long as it is wide. if the distance around the park is 12.5 kilometers, what is the area of the park?
blsea [12.9K]
So, first, what are the two sides?

let's call then x and y

we know that 2(x+y)=12.5 (that's the distance around)

so that means that x+y=6.25 (I just divided both by 2)

now, x=4y (from "4 times as long as it is wide")

so we can substitute:

x+4x=6.25

5x=6.25

x=1.25

so one side, is 1.25 and the other will be 1.25*4=5

and for the area we multiply the two:

1.25*5=6.25 square kilometers, and this is the answer!
7 0
3 years ago
Read 2 more answers
A molasses can is made with a cylindrical base having a radius of 6 cm, that is topped by a smaller cylindrical pouring spout wi
wariber [46]

Answer:

The total height of the can is 20cm

7 0
3 years ago
Please help me out with equations in math!<br> Its a starr review!
Alexeev081 [22]

Answer:

F

Step-by-step explanation:

Multiply -3 by 6.5 and you'll get your answer :)

I hope I helped!

4 0
3 years ago
The total stopping distance T for a vehicle is T = 2.5x + 0.5x 2 , where T is in feet and x is the speed in miles per hour. Appr
Kobotan [32]

Answer:

%  change in stopping distance = 7.34 %

Step-by-step explanation:

The stooping distance is given by

T = 2.5 x + 0.5 x^{2}

We will approximate this distance  using the relation

f (x + dx) = f (x)+ f' (x)dx

dx = 26 - 25 = 1

T' =  2.5 + x

Therefore

f(x)+f'(x)dx = 2.5x+ 0.5x^{2} + 2.5 +x

This is the stopping distance at x = 25

Put x = 25 in above equation

2.5 × (25) + 0.5× 25^{2} + 2.5 + 25 = 402.5 ft

Stopping distance at x = 25

T(25) = 2.5 × (25) + 0.5 × 25^{2}

T(25) = 375 ft

Therefore approximate change in stopping distance = 402.5 - 375 = 27.5 ft

%  change in stopping distance = \frac{27.5}{375} × 100

%  change in stopping distance = 7.34 %

6 0
3 years ago
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