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3 years ago

Please Help!! If answer, please answer them all, thank you!!

Mathematics

1 answer:

Sergio [31]3 years ago

8 0

9514 1404 393

Answer:

(b) angle FOA

(d) angle AOH

Step-by-step explanation:

(b) The rays of vertical angles are opposites that form intersecting lines.

The opposite of ray OG is OF. The opposite of ray OB is OA, so the vertical angle to GOB is angle FOA.

(c) The opposite of ray OB is OA, so the supplement to angle EOB is angle EOA.

(d) Similarly, the supplement to angle BOH is angle AOH.

_____

Comment on supplementary angles

Angles that form a linear pair are supplementary. Angles do not have to form a linear pair to be supplementary. They merely have to have a sum of 180°. Here, the supplementary angles of interest do form a linear pair, so finding the other angle of the pair means only finding the other point that names the line being formed by the pair.

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Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

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Help out?? Mathematics image.

Brilliant_brown [7]

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Step-by-step explanation:

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$\frac{1}{2}$ (UV - 20) = 30

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17) arc AB + arc BC = arc AC

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19) ∠1 + ∠ 2 = arc ABC

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