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3 years ago

Choose and then match the ratio which is equivalent

Mathematics

1 answer:

madreJ [45]3 years ago

4 0

1. 5 to 20 ---> c. 20 to 80

(5/20 and 20/80, both have the same answer(ratio) which is equal to 1/4)

2. 5 to 40 ---> e. 11 to 88

(5/40 and 11/88, both have the same answer(ratio) which is equal to 1/8)

3. 9 to 45 ---> b. 7 to 35

(9/45 and 7/35, both have the same answer(ratio) which is equal to 1/5)

4. 2 to 20 ---> a. 1 to 10

(2/20 and 1/10, both have the same answer(ratio) which is equal to 1/10)

5. 4 to 12 ---> d. 10 to 30

(4/12 and 10/30, both have the same answer(ratio) which is equal to 1/3)

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2 years ago

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Troyanec [42]

f(x) = 4x + 3

That's one-to-one, a linear function that will give a different f(x) for each different x.

For the inverse let's swap x and y and solve for y

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That's the inverse,

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3 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)

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3 years ago