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3 years ago

12

X² + 8x' -2 Someone help me with this ???? Please

1 answer:

quester [9]3 years ago

7 0

Roots: (-4 -3V2 ;0) (-4+3V2 ;0)
Range: y€[-18; +♾)
Minimums: (-4; -18)
Vertical intercept: (0; -2)

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Rewrite the following expression using the properties of rational exponents. Be sure your answer is in simplest form.

Nookie1986 [14]

Answer:

$(2\cdot 6)^{\frac{3}{2}}=24(3)^{\frac{1}{2}}$

Step-by-step explanation:

The given expression is $(2*6)^{\frac{3}{2}}$

We multiply within the parenthesis to get:

$(12)^{\frac{3}{2}}$

Recall that: $a^{\frac{m}{n} } =\sqrt[n]{x^m}$

We apply this property to get:

$(12)^{\frac{3}{2}}=\sqrt{12^3}$

$(12)^{\frac{3}{2}}=\sqrt{12^2*12}$

This simplifies to

$(12)^{\frac{3}{2}}=12\sqrt{12}$

We can simplify further

$(12)^{\frac{3}{2}}=12\sqrt{4*3}=12*2\sqrt{3}=24\sqrt{3}$

We rewrite as rational exponent $24(3)^{\frac{1}{2}}$

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3 years ago

joseph accepted a new job at a company with a contract guaranteeing annual raises. Joseph will get a raise of $5000 every year a

Stels [109]

Answer:S = 70,000 + 5000n

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3 years ago

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If u understand this May u please help me

Sladkaya [172]

4 is the y axis so from the y axis you go up 1 and over right 6!

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3 years ago

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.

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3 years ago

Can a triangle have side lengths 4, 4, 7? <br> 1.no <br> 2. Yes <br> 3. not enough information

Marina CMI [18]

Yes! The lengths of each side must be less than the sum of the other two lengths. It looks like this:
4+4>7
7+4>4
4+7>4

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3 years ago