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4 years ago

Which is equivalent

Mathematics

2 answers:

Citrus2011 [14]4 years ago

7 0

For this case we must find an expression equivalent to:

$(x ^ {\frac {4} {3}} * x ^ {\frac {2} {3}}) ^ {\frac {1} {3}}$

By definition of power properties we have to:

$(a ^ n) ^ m = a ^ {n * m}$

So, rewriting the expression we have:

$x ^ {\frac {4} {3 * 3}} * x ^ {\frac {2} {3 * 3}} =$

$x ^ {\frac {4} {9}} * x ^ {\frac {2} {9}} =$

By definition of multiplication of powers of the same base, we put the same base and add the exponents:

$x ^ {\frac {4} {9} + \frac {2} {9}} =\\x ^ {\frac {4 + 2} {9}} =\\x ^ {\frac {6} {9}} =\\x ^ {\frac {2} {3}}$

Answer:

Option B

Amanda [17]4 years ago

5 0

Answer:

$x^{2/3}$

Step-by-step explanation:

The question is on rules of rational exponents

Here we apply the formulae for product rule where;

$= a^{n} *a^{t} = a^{n+t} \\\\\\\\=(x^{4/3} *x^{2/3} ) = x^{4/3 + 2/3} = x^{6/3} = x^{2} \\\\\\=(x^2)^{1/3} \\\\\\=\sqrt[3]{x^2}$

$=x^{2/3}$

You might be interested in

P=−4b 2 +6b−9 <br> Q=7b 2 −2b−5<br> P−Q=<br> Your answer should be a polynomial in standard form.

shepuryov [24]

Answer:

−11b^2+8b−4

Step-by-step explanation:

−4b2+6b−9−7b2+2b+5

Simplify by adding terms.

11b2+8b−4

3 0

3 years ago

Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

Graph the line with the equation

pochemuha

The graph of the linear equation can be seen in the image below.

<h3 /><h3>How to graph the linear equation?</h3>

Here we have the linear equation:

y = (2/5)*x - 6

To graph it, we just need to find two points on the line, and then connect them with a line.

To find the points we just evaluate in two values of x.

if x = 0.