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jekas [21]
3 years ago
5

the jedi have a weapon that can shoot a projectile 20 miles. find the area that the projectile can land in.

Mathematics
1 answer:
kifflom [539]3 years ago
5 0

Answer:

The area he can cover with that weapon is 1,256 miles².

Step-by-step explanation:

Assuming that the Jedi won't move and that he can shoot in every direction, he'd be shooting at a circle with a radius of 20 miles. Therefore the are that he can cover with that weapon is the area of the circle, which is shown below:

area = pi*r²

area = 3.14*(20)²

area = 3.14*400

area = 1,256 miles²

The area he can cover with that weapon is 1,256 miles².

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The rate would be -2,000 feet per minute.

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Consider the function f(x)=xln(x). Let Tn be the nth degree Taylor approximation of f(2) about x=1. Find: T1, T2, T3. find |R3|
Fynjy0 [20]

Answer:

R3 <= 0.083

Step-by-step explanation:

f(x)=xlnx,

The derivatives are as follows:

f'(x)=1+lnx,

f"(x)=1/x,

f"'(x)=-1/x²

f^(4)(x)=2/x³

Simialrly;

f(1) = 0,

f'(1) = 1,

f"(1) = 1,

f"'(1) = -1,

f^(4)(1) = 2

As such;

T1 = f(1) + f'(1)(x-1)

T1 = 0+1(x-1)

T1 = x - 1

T2 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2

T2 = 0+1(x-1)+1(x-1)^2

T2 = x-1+(x²-2x+1)/2

T2 = x²/2 - 1/2

T3 = f(1)+f'(1)(x-1)+f"(1)/2(x-1)^2+f"'(1)/6(x-1)^3

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Thus, T1(2) = 2 - 1

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T2 (2) = 3/2

T2 (2) = 1.5

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T3(2) = 4/3

T3(2) = 1.333

Since;

f(2) = 2 × ln(2)

f(2) = 2×0.693147 =

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Since;

f(2) >T3; it is significant to posit that T3 is an underestimate of f(2).

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f^(4)(x)=2/x^3, we have, |f^(4)(c)| <= 2

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R3 <= | 2 / 24× 1 |

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