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Maru [420]
3 years ago
6

Please solve!!!!!!!!! Thanks!

Mathematics
1 answer:
Nat2105 [25]3 years ago
4 0

Answer:

  • a) 60°
  • b) 80°
  • c) 100°
  • d) 50°
  • e) 30°

Step-by-step explanation:

The key here is that AB ║ EC. This makes arc AE have the same measure as arc BC. Since those have the same measure as AB and the three arcs together make a semicircle, each has measure 180°/3 = 60°.

Then the various arc measures are:

  • AB = 60°
  • BC = 60°
  • CD = 80° (given)
  • DE = 100° . . . . . since CDE is 180°
  • EA = 60°

Then your answers are ...

a) AE = 60°

b) ∠ABD = (1/2)(DE +EA) = (1/2)(100° +60°) = 80°

c) ∠DFC = (1/2)(CD +EB) = (1/2)(80° + (60° +60°)) = 100°

d) ∠P = (1/2)(DA -AB) = (1/2)(100° +60° -60°) = 50°

e) ∠PAB = (1/2)(AB) = (1/2)(60°) = 30°

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
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Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

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y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

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y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

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y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

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Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

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\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

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