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Solve the initial-value problem y' = x^4 - \frac{1}{x}y, y(1) = 1.

natta225 [31]

The ODE is linear:

$y'=x^4-\dfrac yx$

$y'+\dfrac yx=x^4$

Multiplying both sides by $x$ gives

$xy'+y=x^5$

Notice that the left side can be condensed as the derivative of a product:

$(xy)'=x^5$

Integrating both sides with respect to $x$ yields

$xy=\dfrac{x^6}6+C$

$\implies y(x)=\dfrac{x^5}6+\dfrac Cx$

Since $y(1)=1$ ,

$1=\dfrac16+C\implies C=\dfrac56$

so that

$\boxed{y(x)=\dfrac{x^5}6+\dfrac5{6x}}$

4 0

3 years ago

How do I find x on the attached right triangle ?

ioda

The value of x from the given triangle is 32√3/3

<h3>SOH CAH TOA identity</h3>

In order to determine the value of x, first we need to determine the hypotenuse side of the smaller triangle.

sin 60 = 8√2/H

H = 8√2/sin60

H = 8√2 * 2/√3

H = 16√2/√3
H = 16√6/3

For the value of x, we will use the expression

sin45 = (16√6/3)/x

1/√2 = 16√6/3x
3x = 16√12
3x = 32√3

x = 32√3/3

Hence the value of x from the given triangle is 32√3/3

Learn more on SOH CAH TOA here: brainly.com/question/20734777

#SPJ1

4 0

2 years ago

At a restaurant you only have 20 dollars to spend on dinner. in addition to the cost of the meal you must pay only 5% sales tax

wlad13 [49]

$16.68 is the most expensive item you can order with $20.00

7 0

3 years ago

Find the factors and zeros of 6x^2 +8x-28=2x^2 + 4

Mice21 [21]

Answer:

First solve the equation:

6x^2 + 8x -28 = 2x^2 + 4

=> 6x^2 - 2x^2 + 8x - 28 -4 = 0 => 4x^2 + 8x - 32 = 0

Extract common factor 4:

=> 4[x^2 + 2x - 8] = 0

Now factor the polynomial:

4(x + 4) (x - 2) = 0

=> the solutions are x + 4 = 0 => x = -4, and x - 2 = 0 => x = 2.

So the answer is the option B: 4(x + 4)(x - 2); {-4, 2}

HOPE THIS HELPS! :D

4 0

2 years ago

Calculate the length of sides triangle pqr and determine weather or not triangle is a right angled. P(-4,6) q(6,1) r(2,9)

Tom [10]

$\bold{\huge{\underline{ Solution }}}$

<h3>Given :-</h3>

We have given the coordinates of the triangle PQR that is P(-4,6) , Q(6,1) and R(2,9)

We have to calculate the length of the sides of given triangle and also we have to determine whether it is right angled triangle or not

<h3>Let's Begin :-</h3>

Here, we have 

Coordinates of P =( x1 = -4 , y1 = 6)
Coordinates of Q = ( x2 = 6 , y2 = 1 )
Coordinates of R = ( x3 = 2 , y3 = 9 )

By using distance formula 

$\pink{\bigstar}\boxed{\sf{Distance=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2\;}}}$

Subsitute the required values in the above formula :-

Length of side PQ

$\sf{ = }{\sf\sqrt{ (6 - (-4))^{2} + (1 - 6)^{2}}}$

$\sf{ = }{\sf\sqrt{ (6 + 4 )^{2} + (- 5)^{2}}}$

$\sf{ = }{\sf\sqrt{ (10)^{2} + (- 5)^{2}}}$

$\sf{ = }{\sf\sqrt{ 100 + 25 }}$

$\sf{ = }{\sf\sqrt{ 125 }}$

$\sf{ = 5 }{\sf\sqrt{ 5 }}$

Length of QR

$\sf{ = }{\sf\sqrt{(2 - 6)^{2} + (9 - 1)^{2}}}$

$\sf{ = }{\sf\sqrt{(- 4 )^{2} + (8)^{2}}}$

$\sf{ = }{\sf\sqrt{16 + 64 }}$

$\sf{ = }{\sf\sqrt{80 }}$

$\sf{ = 4 }{\sf\sqrt{5 }}$

Length of RP

$\sf{ = }{\sf\sqrt{ (-4 - 2 )^{2} + (6 - 9)^{2}}}$

$\sf{ = }{\sf\sqrt{ (-6 )^{2} + (-3)^{2}}}$

$\sf{ = }{\sf\sqrt{ 36 + 9 }}$

$\sf{ = }{\sf\sqrt{ 45 }}$

$\sf{ = 3}{\sf\sqrt{ 5 }}$

We have to determine whether the triangle PQR is right angled triangle

<h3>Therefore, </h3>

By using Pythagoras theorem :-

Pythagoras theorem states that the sum of squares of two sides that is sum of squares of 2 smaller sides of triangle is equal to the square of hypotenuse that is square of longest side of triangle

That is,

$\bold{ PQ^{2} + QR^{2} = PR^{2}}$

Subsitute the required values,

$\bold{ 125 + 80 = 45 }$

$\bold{ 205 = 45 }$

From above we can conclude that,

The triangle PQR is not a right angled triangle because 205 ≠ 45 .

6 0

2 years ago