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3 years ago

Adjacent, right angles are complementary. always sometimes never

2 answers:

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:

solmaris [256]

Answer:

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

4 0

3 years ago

Read 2 more answers

Find the perimeter of the figure to the nearest hundredth.

bekas [8.4K]

Answer:

38.56

Step-by-step explanation:

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2 years ago

John is saving money to buy a game. so far he has saved $12 , which is one-fourth of the total cost of the game. how much does t

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One fourth of the game cost $3

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3 years ago

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hare been better is he had velict

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2 years ago

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

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

as desired. So one possible ODE would be

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

(See "Euler-Cauchy equation" for more info)

6 0

3 years ago