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2 years ago

Uhbjhbjhbjhbjhbjhbjhbfg

Mathematics

1 answer:

Len [333]2 years ago

5 0

Answer:

2/5 (B)

Step-by-step explanation:

$x^\frac{1}{y}=\sqrt[y]{x}$

So:

$\sqrt{\frac{4}{25}}=\frac{2}{5}$ because

$(\frac{x}{y})^z=\frac{x^z}{y^z}$

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2 11/12 cups of flour

Step-by-step explanation:

2 recipes are 2 1/2 cups of flour

1/3 is 1/3 x 1 1/4= 1/3 + 1/12= 5/12

So, 2 11/12 cups of flour

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Step-by-step explanation:

28/140=0.2

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2 years ago

Read 2 more answers

Solve Quadratic Equations by Factoring:

mr_godi [17]

Answer:

x = 0 or 1
x = ±i/4
x = -5 (twice)

Step-by-step explanation:

Factoring is aided by having the equations in standard form. The first step in each case is to put the equations in that form. The zero product property tells you that a product is zero when a factor is zero. The solutions are the values of x that make the factors zero.

1. x^2 -x = 0

x(x -1) = 0 . . . . . x = 0 or 1

2. 16x^2 +1 = 0

This is the "difference of squares" ...

(4x)^2 - (i)^2 = 0

(4x -i)(4x +i) = 0 . . . . . x = -i/4 or i/4 (zeros are complex)

3. x^2 +10x +25 = 0

(x +5)(x +5) = 0 . . . . . x = -5 with multiplicity 2

5 0

2 years ago

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

2 years ago