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Bond [772]
3 years ago
7

Adjacent, right angles are complementary. always sometimes never

Mathematics
2 answers:
Harrizon [31]3 years ago
8 0
B. <span>sometimes
is the right answer
</span> <span>Any two angles which add up to 90 degrees are complementary.</span>
Mandarinka [93]3 years ago
4 0

Answer:

the answer is never trust me i had this on a test and got it right




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The least common denominator of 1/2, 1/6, and 1/9 is:
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Answer:

18

Step-by-step explanation:

Find the factors of each denominator

2 = 1*2

6 = 2*3

9 = 3*3

The least common multiple is

2*3*3 = 18

The least common denominator is 18

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Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu
Alisiya [41]
Let y=C_1x+C_2x^3=C_1y_1+C_2y_2. Then y_1 and y_2 are two fundamental, linearly independent solution that satisfy

f(x,y_1,{y_1}',{y_1}'')=0
f(x,y_2,{y_2}',{y_2}'')=0

Note that {y_1}'=1, so that x{y_1}'-y_1=0. Adding y'' doesn't change this, since {y_1}''=0.

So if we suppose

f(x,y,y',y'')=y''+xy'-y=0

then substituting y=y_2 would give

6x+x(3x^2)-x^3=6x+2x^3\neq0

To make sure everything cancels out, multiply the second degree term by -\dfrac{x^2}3, so that

f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y

Then if y=y_1+y_2, we get

-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0

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-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0

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6 0
3 years ago
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