All the factors of 21

Solve the equation using the substitution u = y/x. When u = y/x is substituted into the equation, the equation becomes separable

bekas [8.4K]

Answer:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

Step-by-step explanation:

First step: I'm going to solve our substitution for y:

$u=\frac{y}{x}$

Multiply both sides by x:

$ux=y$

Second step: Differentiate the substitution:

$u'x+u=y'$

Third step: Plug in first and second step into the given equation dy/dx=f(x,y):

$u'x+u=\frac{x(ux)+(ux)^2}{3x^2+(ux)^2}$

$u'x+u=\frac{ux^2+u^2x^2}{3x^2+u^2x^2}$

We are going to simplify what we can.

Every term in the fraction on the right hand side of equation contains a factor of $x^2$ so I'm going to divide top and bottom by $x^2$ :

$u'x+u=\frac{u+u^2}{3+u^2}$

Now I have no idea what your left hand side is suppose to look like but I'm going to keep going here:

Subtract u on both sides:

$u'x=\frac{u+u^2}{3+u^2}-u$

Find a common denominator: Multiply second term on right hand side by $\frac{3+u^2}{3+u^2}$ :

$u'x=\frac{u+u^2}{3+u^2}-\frac{u(3+u^2)}{3+u^2}$

Combine fractions while also distributing u to terms in ( ):

$u'x=\frac{u+u^2-3u-u^3}{3+u^2}$

$u'x=\frac{-u^3+u^2-2u}{3+u^2}$

Third step: I'm going to separate the variables:

Multiply both sides by the reciprocal of the right hand side fraction.

$u' \frac{3+u^2}{-u^3+u^2-2u}x=1$

Divide both sides by x:

$\frac{3+u^2}{-u^3+u^2-2u}u'=\frac{1}{x}$

Reorder the top a little of left hand side using the commutative property for addition:

$\frac{u^2+3}{-u^3+u^2-2u}u'=\frac{1}{x}$

The expression on left hand side almost matches your expression but not quite so something seems a little off.

5 0

3 years ago

Is 16:14 in 64:60 eqiluvant

Elanso [62]

16: 14 - divide by 2

16/2= 8 , 14/2=7

64:60- divide by 4

64/4= 16 , 60/4= 15

Answer: 16:14 and 64:60 are not equivalent.

4 0

3 years ago

Measurements of regular hexagon

TEA [102]

The regular hexagon has both reflection symmetry and rotation symmetry.

Reflection symmetry is present when a figure has one or more lines of symmetry. A regular hexagon has 6 lines of symmetry. It has a 6-fold rotation axis.

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Rotation symmetry is present when a figure can be rotated (less than 360°) and still look the same as before it was rotated. The center of rotation is a point a figure is rotated around such that the rotation symmetry holds. A regular hexagon can be rotated 6 times at an angle of 60°

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5 0

3 years ago

write a paragraph summary of the information provided in the videos. Your summary should include information on the coordinate s

sleet_krkn [62]

How is anyone supposed to help without the video?

8 0

4 years ago

Read 2 more answers

How many and of which kind of roots does the equation

Yakvenalex [24]

Answer:

D. 1 real; 4 complex

Step-by-step explanation:

1. Calculate the total number of roots

The fundamental theorem of algebra states that any polynomial of degree n has n roots.

Your polynomial is fifth degree, so it has five roots.

2. Calculate the number of real roots

We can use Descartes' rule of signs to determine the number of real roots:

The number of positive real roots is the same as the number of changes in the sign of the coefficients of ƒ(x) or less than by an even number.

The number of negative real roots is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.

(a) Number of positive real roots

The coefficients of ƒ(x) are

+1 +3 +8 +14 +16 + 8

There are no changes of sign, so there are no positive real roots.

(b) Number of negative real roots

(i) Find f(-x)

f(-x) = -x⁵+ 3x⁴ - 8x³ + 14x² - 16x + 8

(ii) Count the changes of sign

The coefficients of f(-x) are

-1 ∥ +3 ∥ -8 ∥+14 ∥ -16 ∥ +8

There are five changes of sign.

Thus, the number of negative roots is five, three, or one.

So far, we know that we have five roots. None is positive, and there could be one, three, or five negative real roots.

3. Confirm by sketching the graph

The real roots are the points at which the graph crosses or touches the x-axis.

The graph (see below) crosses the x-axis at only one point.

Thus. we have

one real root (negative)
four complex roots

3 0

3 years ago