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3 years ago

A and b are positive integers and a–b = 3. Evaluate the following: 125^(1/3a)/25^(1/2b)

Mathematics

2 answers:

Snezhnost [94]3 years ago

5 0

Answer:

125

Step-by-step explanation:

$\dfrac{125^{\frac{1}{3}a}}{25^{\frac{1}{2}b}} =$

$= \dfrac{(5^3)^{\frac{1}{3}a}}{(5^2)^{\frac{1}{2}b}}$

$= \dfrac{5^{3 \times \frac{1}{3}a} }{5^{2 \times \frac{1}{2}b}}$

$= \dfrac{5^{\frac{3}{3}a}}{5^{\frac{2}{2}b}}$

$= \dfrac{5^a}{5^b}$

$= 5^{a - b}$

$= 5^3$

$= 125$

gizmo_the_mogwai [7]3 years ago

4 0

Value of $\dfrac{125^{1/3a}}{25^{1/2b}}$ is $\dfrac{1}{125^{(\frac{1}{ab})}}$ .

<u>Step-by-step explanation:</u>

Here we have , a-b=3 . We need to evaluate : 125^(1/3a)/25^(1/2b) or ,

$\dfrac{125^{1/3a}}{25^{1/2b}}$ . Let's find out:

⇒ $\dfrac{125^{1/3a}}{25^{1/2b}}$

⇒ $\dfrac{5^3(^{1/3a})}{5^2(^{1/2b})}$

⇒ $\dfrac{5(^{3/3a})}{5(^{3/2b})}$

⇒ $5^{(\frac{1}{a}-\frac{1}{b})} = 5^{(\frac{b-a}{ab})}$

⇒ $5^{(\frac{-3}{ab})}$

⇒ $\dfrac{1}{125^{(\frac{1}{ab})}}$

Therefore, Value of $\dfrac{125^{1/3a}}{25^{1/2b}}$ is $\dfrac{1}{125^{(\frac{1}{ab})}}$ .

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b. Rate at which the revenue changing in the year 2010 is 76 million dollar per year ( decreasing)

Step-by-step explanation:

For part a,

Given that x is the number of years at the beginning of 2007.

Therefore, $x=0$ for year 2007.

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Substitute the value of x=3 in R(x),

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$R\left(x\right)=-75$

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Therefore, there is loss of revenue at the beginning of 2010 which is 75 million dollar.

For part b,

To calculate rate, differentiate the given function with respect to x.

$\dfrac{d}{dx}R\left(x\right)=\dfrac{d}{dx}\left (-\left(x\right)^{4}+8\left(x\right)^{3}-38\left(x\right)^{2}+44\left(x\right)\right)$

Applying sum and difference rule of derivative,

$\dfrac{d}{dx}R\left(x\right)=-\dfrac{d}{dx}\left(x^4\right)+\dfrac{d}{dx}\left(8x^3\right)-\dfrac{d}{dx}\left(38x^2\right)+\dfrac{d}{dx}\left(44x\right)$

Applying constant multiple rule of derivative,

$\dfrac{d}{dx}R\left(x\right)=-\dfrac{d}{dx}\left(x^4\right)+8\dfrac{d}{dx}\left(x^3\right)-38\dfrac{d}{dx}\left(x^2\right)+44\dfrac{d}{dx}\left(x\right)$

Applying power rule of derivative,

$\dfrac{d}{dx}R\left(x\right)=-\left(4x^{4-1}\right)+8\left(3x^{3-1}\right)-38\left(2x^{2-1}\right)+44\left(1x^{1-1}\right)$

$\dfrac{d}{dx}R\left(x\right)=-4\left(x^{3}\right)+8\left(3x^{2}\right)-38\left(2x^{1}\right)+44$

$\dfrac{d}{dx}R\left(x\right)=-4x^3+24x^2-76x+44$

Substituting the value x=3,

$\dfrac{d}{dx}R\left(x\right)=-4\left(3\right)^3+24\left(3\right)^2-76\left(3\right)+44$

$\dfrac{d}{dx}R\left(x\right)=-4\left(27\right)+24\left(9\right)-76\left(3\right)+44$

$\dfrac{d}{dx}R\left(x\right)=-108+216-228+44$

$\dfrac{d}{dx}R\left(x\right)=-76$

Negative sign indicates that rate is decreasing.

Rate at which the revenue is changing in the year 2010 is 76 million dollar per year.

8 0

3 years ago