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charle [14.2K]
3 years ago
15

On a trip to the museum, lance saw a replica of a circular shield used by ancient greek soldiers. if the circumference of the sh

ield is 100.48 inches, its radius is inches and the area of the front of the shield is about square inches.
Mathematics
2 answers:
SOVA2 [1]3 years ago
6 0
If the circumference of the shield is 100.48inches, nits radius is 16 inches and the area of the front of the shield is about 803.84 square inches.
proof

circumference = 2x r x Pi implies r =<span>circumference / 2Pi=100.48/2x3.14=16
area = Pi x r²=3.14 x 16²=804.84

</span>
adoni [48]3 years ago
5 0

Answer:

The radius of shield is 16 inches and the area of front of the shield is 803.84 inches^2

Step-by-step explanation:

Circumference of circle = 2\pi r

Circumference of the shield is 100.48 inches

⇒2 \pi r = 100.48

⇒2\times 3.14 \times r = 100.48

⇒6.28\times r = 100.48

⇒r = \frac{100.48}{6.28}

⇒r = 16

Thus the radius of shield is 16 inches

Area of shield = \pi r^2

                       = 3.14 \times (16)^2

                       = 803.84 inches^2

Hence the radius of shield is 16 inches and the area of front of the shield is 803.84 inches^2

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Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de
Ganezh [65]

Answer:

a. \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

b. \mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

Step-by-step explanation:

The initial value problem is given as:

y' +y = 7+\delta (t-3) \\ \\ y(0)=0

Applying  laplace transformation on the expression y' +y = 7+\delta (t-3)

to get  L[{y+y'} ]= L[{7 + \delta (t-3)}]

l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}

Taking inverse of Laplace transformation

y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{e^{-3s}}{s+1}]

L^{-1}[\dfrac{1}{s+1}] = e^{-t}  = f(t) \ then \ by \ second \ shifting \ theorem;

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \  \ \  \ t

= e^{-t-3} \left \{ {{1 \ \ \ \ \  t>3} \atop {0 \ \ \ \ \  t

= e^{-(t-3)} u (t-3)

Recall that:

y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]

Then

y(t) = 7 -7e^{-t}  +e^{-(t-3)} u (t-3)

y(t) = 7 -7e^{-t}  +e^{-t} e^{-3} u (t-3)

\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}

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Sati [7]
The other endpoint (x,y)
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coldgirl [10]

Answer:

I think the answer is B

Step-by-step explanation:

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\bold{ANSWER:}
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