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Nonamiya [84]
3 years ago
10

3. The fourth root of x.

Mathematics
1 answer:
Hunter-Best [27]3 years ago
6 0

Answer:

It depends what x equals.

Step-by-step explanation:

There's plenty of resources on places like Khan Academy to help.

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Find the general solution of the differential equation. (Use C for any needed constant.) dy dx = 1 − y
daser333 [38]

Answer:

Step-by-step explanation:

Give the DE

dy/dx = 1-y

Using variable separable method

dy = (1-y)dx

dx = dy/(1-y)

Integrate both sides

 ∫dx =  ∫dy/(1-y)

 ∫dy/(1-y)=  ∫dx

-ln(1-y) = x+C

ln(1-y)^-1 = x+C

Apply e to both sides

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4 0
4 years ago
1. **Which of the foll byowing is equivalent to y4 x y^8?
andrey2020 [161]

Answer:

<h2><em><u>S</u></em><em><u>O</u></em><em><u>L</u></em><em><u>U</u></em><em><u>T</u></em><em><u>I</u></em><em><u>O</u></em><em><u>N</u></em></h2>

{y}^{4}  \times  {y}^{8}  = y8 + 4 =  {y}^{12}

<h3><em>B</em><em>e</em><em>a</em><em>c</em><em>a</em><em>u</em><em>s</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>p</em><em>r</em><em>o</em><em>p</em><em>e</em><em>r</em><em>t</em><em>y</em><em> </em><em>o</em><em>f</em><em> </em><em>e</em><em>x</em><em>p</em><em>o</em><em>n</em><em>e</em><em>n</em><em>t</em><em> </em><em>=</em><em> </em></h3><h3><em>a ^{b}  \times  {a}^{c}  = a ^{b + c}</em></h3>
8 0
3 years ago
What is the solution to the linear equation? x – StartFraction 2 Over 3 EndFraction x minus StartFraction one-half EndFraction e
Anton [14]

<em>Your question is not well presented</em>

Question:

What is the solution to the linear equation?

x - \frac{2}{3}x - \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x

Answer:

x = \frac{-5}{3}

Step-by-step explanation:

Given

x - \frac{2}{3}x - \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x

Required

Simplify

x - \frac{2}{3}x - \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x

Add \frac{1}{2} to both sides

x - \frac{2}{3}x - \frac{1}{2} + \frac{1}{2} = \frac{1}{3} + \frac{5}{6}x + \frac{1}{2}

x - \frac{2}{3}x = \frac{1}{3} + \frac{5}{6}x + \frac{1}{2}

Subtract \frac{5}{6}x from both sides

x - \frac{2}{3}x - \frac{5}{6}x= \frac{1}{3} + \frac{5}{6}x + \frac{1}{2} - \frac{5}{6}x

x - \frac{2}{3}x - \frac{5}{6}x= \frac{1}{3} + \frac{1}{2} + \frac{5}{6}x- \frac{5}{6}x

x - \frac{2}{3}x - \frac{5}{6}x= \frac{1}{3} + \frac{1}{2}

Take LCMs

\frac{6x - 4x -5x}{6}= \frac{2+3}{6}

\frac{-3x}{6}= \frac{5}{6}

Multiply both sides by 6

6 * \frac{-3x}{6}= \frac{5}{6} * 6

-3x= 5

Divide both sides by -3

\frac{-3x}{-3}= \frac{5}{-3}

x = \frac{5}{-3}

x = \frac{-5}{3}

5 0
4 years ago
Read 2 more answers
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