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EastWind [94]
3 years ago
8

What is the expression in simplest radical form?

rac%7B2%7D%7B9%7D%20%7D" id="TexFormula1" title="(5x^{4} y^{3})^{\frac{2}{9} }" alt="(5x^{4} y^{3})^{\frac{2}{9} }" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
Vlad [161]3 years ago
5 0

Use the formula a^(x/n) = (n)√a^x   (note it is a small n)

(5x^4y^3)^(2/9) = Small 9

Convert.

\sqrt{9{{(25x^{8})y^9 }}} is your answer

hope this helps

sergey [27]3 years ago
5 0

So here are a few things about exponents you should know:

  1. Fractional exponents to radicals: x^\frac{m}{n}=\sqrt[n]{x^m}
  2. Powering a power: (x^m)^n=x^{m*n}

So firstly, convert the fractional exponent to a radical:

(5x^4y^3)^\frac{2}{9}=\sqrt[9]{(5x^4y^3)^2}

Next, solve the outer power:

\sqrt[9]{(5x^4y^3)^2} =\sqrt[9]{5^2x^{4*2}y^{3*2}} =\sqrt[9]{25x^8y^6}

Your final answer is \sqrt[9]{25x^8y^6}

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What are the approximate values of the minimum and maximum points of f(x) = x5 − 10x3 + 9x on [-3,3]?
nika2105 [10]

Answer:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.

Step-by-step explanation:

Given:

f(x)=x^5-10x^3+9x; [-3,3]

Explanation:

In order to find minimum/maximum of a function, we need to find the first derivative of the function and then set it equal to 0 to get critical points.

Therefore,

f'(x)=5x^4-30x^2+9

Setting derivative equal to 0, we get

5x^4-30x^2+9=0

On applying quadratic formula, we get

x=2.4, -2.4, -0.7, 0.7.

So, those are critical points of the given function.

Plugging the values x=2.4, -2.4, -0.7, 0.7, -3 and 3 in above function, we get

f(2.4)=(2.4)^5-10(2.4)^3+9(2.4)= -37.01376   : Minimum.

f(-2.4)=(-2.4)^5-10(-2.4)^3+9(-2.4)= 37.01376 : Maximum.

f(0.7)=(0.7)^5-10(0.7)^3+9(0.7) = 3.03807

f(-0.7)=(-0.7)^5-10(-0.7)^3+9(-0.7) = -3.03807

f(-3)=(-3)^5-10(-3)^3+9(-3) =0

f(3)=(3)^5-10(3)^3+9(3) =0

Therefore the approximate values of the minimum and maximum points of f(x) = x^5- 10x^3+ 9x on [-3,3] are:

Minimum : -37 at x=2.4 and

Maximum = 37 at x=-2.4.


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