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Alina [70]
3 years ago
5

A rug shaped like a rectangle has a width of 3 m. the length of the rug is 2 m greater than its width. what is the perimeter of

the rug in meters
Mathematics
1 answer:
Wewaii [24]3 years ago
3 0
Well,
rectangle has 4 sides
2 lengths and 2 breadths
the width is the breadth
so it's 3x2= 6m
the leanght
it's 2m greater than width so it's (3+2)x2= 10m
so the perimeter is the total length of all sides therefore it's 10+6= 16m

it's pretty easy one could just
(3x2)+((3+2)2)= 16m
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1/2 = 2/4 = 3/6 = 4/8 = 5/10 = 6/12

Let's think of this visually, as a pie. 
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AD and MN are chords that intersect at point B.
vampirchik [111]

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4 units

Step-by-step explanation:

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On day t=0t=0t, equals, 0, the stock is at its average value of {\$}3.47$3.47dollar sign, 3, point, 47 per share, but 91.2591.25
Norma-Jean [14]

Answer:

S(t) = a.sin (b.t) + d

a = -1.5, b = (2π/365), d = 3.47

S(t) = -1.5 sin (2πt/365) + 3.47

Step-by-step explanation:

Complete Question is presented in the attached image to this solution.

- Dingane has been observing a certain stock for the last few years and he sees that it can be modeled as a function S(t) of time t (in days) using a sinusoidal expression of the form

S(t) = a.sin(b.t) + d.

On day t = 0, the stock is at its average value of $3.47 per share, but 91.25 days later, its value is down to its minimum of $1.97.

Find S(t). t should be in radians.

S(t) =

Solution

S(t) = a.sin(b.t) + d.

At t = 0, S(t) = $3.47

S(0) = a.sin(b×0) + d = a.sin 0 + d = 3.47

Sin 0 = 0,

S(t=0) = d = 3.47.

At t = 91.25 days, S(t) = $1.97

But, it is given that T has to be in radians, for t to be in radians, the constant b has to convert t in days to radians.

Hence, b = (2π/365)

S(91.25) = 1.97 = a.sin(b×91.25) + d

d = 3.47 from the first expression

S(t = 91.25) = a.sin (91.25b) + 3.47 = 1.97

1.97 = a.sin (2π×91.25/365) + 3.47

1.97 = a sin (0.5π) + 3.47

Sin 0.5π = 1

1.97 = a + 3.47

a = -1.5

Hence,

S(t) = a.sin (b.t) + d

a = -1.5, b = (2π/365), d = 3.47

S(t) = -1.5 sin (2πt/365) + 3.47

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3 years ago
Assume that f(x)=ln(1+x) is the given function and that Pn represents the nth Taylor Polynomial centered at x=0. Find the least
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Answer:

the least integer for n is 2

Step-by-step explanation:

We are given;

f(x) = ln(1+x)

centered at x=0

Pn(0.2)

Error < 0.01

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[[Max(f^(n+1) (c))]/(n + 1)!] × 0.2^(n+1) < 0.01

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First derivative: f'(x) = 1/(x + 1) < 0! = 1

2nd derivative: f"(x) = -1/(x + 1)² < 1! = 1

3rd derivative: f"'(x) = 2/(x + 1)³ < 2! = 2

4th derivative: f""(x) = -6/(x + 1)⁴ < 3! = 6

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Let's try n = 2

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This is less than 0.01.

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