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kati45 [8]
3 years ago
13

How do you know a radical expression is in simplest form?

Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
7 0
Answer: To know whether a radical expression is in simplest form or not you should put the numbers and letters inside the radical in terms of prime factors. Then, the radical expression is in the simplest form if all the numbers and letters inside the radical are prime factors with a power less than the index of the radical

Explanation:

Any prime factor raised to a power greater than the index of the root can be simplified and any factor raised to a power less than the index of the root cannot be simplified

For example simplify the following radical in its simplest form:


\sqrt[5]{3645 a^8b^7c^3}

1) Factor 3645 in its prime factors: 3645 = 3^6 * 5

2) Since the powr of 3 is 6, and  6 can be divided by the index of the root, 5, you can simplify in this way:

- 6 ÷ 5 = 1 with reminder 1, so 3^1 leaves the radical and 3^1 stays in the radical

3) since the factor 5 has power 1 it can not leave the radical

4) the power of a is 8, then:

8 ÷ 5 = 1 with reminder 3 => a^1 leaves the radical and a^3 stays inside the radical.

5) the power of b is 7, then:

7 ÷ 5 = 1 with reminder 2 => b^1 leaves the radical and b^2 stays inside the radical

6) the power of c is 3. Since 3 is less than 5 (the index of the radical) c^3 stays inside the radical.

7) the expression simplified to its simplest form is

3ab \sqrt[5]{3.5.a^3b^2c^3}

And you know it cannot be further simplified because all the numbers and letters inside the radical are prime factors with a power less than the index of the radical.
kow [346]3 years ago
6 0
To simplify, the trick is to split the number into factors where one is a perfect square. Expressing in simplest radical form<span> just means simplifying a </span>radical<span> so that there are no more square roots, cube roots, 4th roots, etc left to find. It also means removing any </span>radicals<span> in the denominator of a fraction.</span>
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Find all possible values of the expression 1/y if: .125
34kurt

Answer:

The possible values for 1/y of the expression 0.125 < y < 0.25 are in the range

4 < 1/y < 8

Step-by-step explanation:

The given information are;

The range of values of y are 0.125 < y < 0.25, therefore, we have;

The boundaries of the function, y are 0.125 and 0.25

The inverse of the boundaries of the function, y are 1/0.125 = 8 and 1/0.24 = 4

Therefore;

The limits of the inverse of the function y are ;

The inequality that represents 1/y is therefore;

1/0.25 < 1/y < 1/0.125 or 4 < 1/y < 8

The possible values of 1/y for the expression 0.125 < y < 0.25 are therefore;

4 < 1/y < 8.

8 0
3 years ago
All boxes with a square​ base, an open​ top, and a volume of 60 ft cubed have a surface area given by ​S(x)equalsx squared plus
Karo-lina-s [1.5K]

Answer:

The absolute minimum of the surface area function on the interval (0,\infty) is S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

The dimensions of the box with minimum surface​ area are: the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

Step-by-step explanation:

We are given the surface area of a box S(x)=x^2+\frac{240}{x} where x is the length of the sides of the base.

Our goal is to find the absolute minimum of the the surface area function on the interval (0,\infty) and the dimensions of the box with minimum surface​ area.

1. To find the absolute minimum you must find the derivative of the surface area (S'(x)) and find the critical points of the derivative (S'(x)=0).

\frac{d}{dx} S(x)=\frac{d}{dx}(x^2+\frac{240}{x})\\\\\frac{d}{dx} S(x)=\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(\frac{240}{x}\right)\\\\S'(x)=2x-\frac{240}{x^2}

Next,

2x-\frac{240}{x^2}=0\\2xx^2-\frac{240}{x^2}x^2=0\cdot \:x^2\\2x^3-240=0\\x^3=120

There is a undefined solution x=0 and a real solution x=2\sqrt[3]{15}. These point divide the number line into two intervals (0,2\sqrt[3]{15}) and (2\sqrt[3]{15}, \infty)

Evaluate S'(x) at each interval to see if it's positive or negative on that interval.

\begin{array}{cccc}Interval&x-value&S'(x)&Verdict\\(0,2\sqrt[3]{15}) &2&-56&decreasing\\(2\sqrt[3]{15}, \infty)&6&\frac{16}{3}&increasing \end{array}

An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We can see from the table that f(x) decreases before x=2\sqrt[3]{15}, increases after it, and is defined at x=2\sqrt[3]{15}. So f(x) has a relative minimum point at x=2\sqrt[3]{15}.

To confirm that this is the point of an absolute minimum we need to find the second derivative of the surface area and show that is positive for x=2\sqrt[3]{15}.

\frac{d}{dx} S'(x)=\frac{d}{dx}(2x-\frac{240}{x^2})\\\\S''(x) =\frac{d}{dx}\left(2x\right)-\frac{d}{dx}\left(\frac{240}{x^2}\right)\\\\S''(x) =2+\frac{480}{x^3}

and for x=2\sqrt[3]{15} we get:

2+\frac{480}{\left(2\sqrt[3]{15}\right)^3}\\\\\frac{480}{\left(2\sqrt[3]{15}\right)^3}=2^2\\\\2+4=6>0

Therefore S(x) has a minimum at x=2\sqrt[3]{15} which is:

S(2\sqrt[3]{15})=(2\sqrt[3]{15})^2+\frac{240}{2\sqrt[3]{15}} \\\\2^2\cdot \:15^{\frac{2}{3}}+2^3\cdot \:15^{\frac{2}{3}}\\\\4\cdot \:15^{\frac{2}{3}}+8\cdot \:15^{\frac{2}{3}}\\\\S(2\sqrt[3]{15})=12\cdot \:15^{\frac{2}{3}} \:ft^2

2. To find the third dimension of the box with minimum surface​ area:

We know that the volume is 60 ft^3 and the volume of a box with a square base is V=x^2h, we solve for h

h=\frac{V}{x^2}

Substituting V = 60 ft^3 and x=2\sqrt[3]{15}

h=\frac{60}{(2\sqrt[3]{15})^2}\\\\h=\frac{60}{2^2\cdot \:15^{\frac{2}{3}}}\\\\h=\sqrt[3]{15} \:ft

The dimension are the base edge x=2\sqrt[3]{15}\:ft and the height h=\sqrt[3]{15} \:ft

6 0
3 years ago
A rectangular board is 1.6 meters long and 1.2 meters wide. What is the area of the board in square millimeters? Do not round yo
Arisa [49]
<h2>Area of rectangular board is 1920000 square millimeters</h2>

Step-by-step explanation:

A rectangular board is 1.6 meters long and 1.2 meters wide.

Here we need to find the area of the board in square millimeters.

Length = 1.6 m= 1.6 x 1000 = 1600 mm

Width = 1.2 m= 1.2 x 1000 = 1200 mm

Area = Length x Width

Area = 1600 x 1200

Area = 1920000 square millimeters

Area of rectangular board is 1920000 square millimeters

6 0
4 years ago
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