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SpyIntel [72]
3 years ago
7

Which expression is it equivalent to?

Mathematics
1 answer:
horrorfan [7]3 years ago
8 0
Option A) Is the answer. \boxed{\mathbf{\dfrac{3f^3}{g^2}}}

For this question; You are needed to expose yourselves to popular usages of radical rules. In this we distribute the squares as one-and-a-half fractions as the squares eliminate the square roots. So, as per the use of fraction conversion from roots. It becomes relatively easy to solve and finish the whole process more quicker than everyone else. More easier to remember.

Starting this with the equation editor interpreter for mathematical expressions, LaTeX. Use of different radical rules will be mentioned in between the steps.

Radical equation provided in this query.

\mathbf{\sqrt{\dfrac{900f^6}{100g^4}}}

Divide the numbered values of 900 and 100 by cancelling the zeroes to get "9" as the final product in the next step.

\mathbf{\sqrt{\dfrac{9f^6}{g^4}}}

Imply and demonstrate the rule of radicals. In this context we will use the radical rule for fractions in which a fraction with a denominator of variable "a" representing a number or a variable, and the denominator of variable "b" representing a number or a variable are square rooted by a value of "n" where it can be a number, variable, etc. Here, the radical of "n" is distributed into the denominator as well as the numerator. Presuming the value of variable "a" and "b" to be greater than or equal to the value of zero. So, by mathematical expression it becomes:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{\dfrac{a}{b}} = \dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}, \: \: a \geq 0 \: \: \: b \geq 0}}

\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{\sqrt{g^4}}}

Apply the radical exponential rule. Here, the squar rooted value of radical "n" is enclosing another variable of "a" which is raised to a power of another variable of "m", all of them can represent numbers, variables, etc. They are then converted to a fractional power, that is, they are raised to an exponent as a fractional value with variables constituting "m" and "n", for numerator and denominator places, respectively. So:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}

\mathbf{Since, \quad \sqrt{g^4} = g^{\frac{4}{2}}}

\mathbf{\therefore \quad \dfrac{\sqrt{9f^6}}{g^2}}

Exhibit the radical rule for two given variables in this current step to separate the variable values into two new squares of variables "a" and "b" with a radical value of "n". Variables "a" and "b" being greater than or equal to zero.

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}, \: \: a \geq 0 \: \: \: b \geq 0}}

So, the square roots are separated into root of 9 and a root of variable of "f" raised to the value of "6".

\mathbf{\therefore \quad \dfrac{\sqrt{9} \sqrt{f^6}}{g^2}}

Just factor out the value of "3" as 3 × 3 and join them to a raised exponent as they are having are similar Base of "3", hence, powered to a value of "2".

\mathbf{\therefore \quad \dfrac{\sqrt{3^2} \sqrt{f^6}}{g^2}}

The radical value of square root is similar to that of the exponent variable term inside the rooted enclosement. That is, similar exponential values. We apply the following radical rule for these cases for a radical value of variable "n" and an exponential value of "n" with a variable that is powered to it.

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^n} = a^{\frac{n}{n}} = a}}

\mathbf{\therefore \quad \dfrac{3 \sqrt{f^6}}{g^2}}

Again, Apply the radical exponential rule. Here, the squar rooted value of radical "n" is enclosing another variable of "a" which is raised to a power of another variable of "m", all of them can represent numbers, variables, etc. They are then converted to a fractional power, that is, they are raised to an exponent as a fractional value with variables constituting "m" and "n", for numerator and denominator places, respectively. So:

\boxed{\mathbf{Radical \: \: Rule: \sqrt[n]{a^m} = a^{\frac{m}{n}}, \: \: a \geq 0}}

\mathbf{Since, \quad \sqrt{f^6} = f^{\frac{6}{2}} = f^3}

\boxed{\mathbf{\underline{\therefore \quad Required \: \: Answer: \dfrac{3f^3}{g^2}}}}

Hope it helps.
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which is the equation of a line that passes through the point (8, 2) and is perpendicular to the line y = -1/4x + 2? A y = -1/4x
Ymorist [56]

Answer: C

Step-by-step explanation:

If a line is perpen to another line, then the other line has the opposite reciprocal of the slope.

y=-(1/4)x+2

slope of the other equation is +4

oly answer with positive slope is C

4 0
3 years ago
What correctly describes the graph. Plsssss help I don’t want to fail
BlackZzzverrR [31]

It could be c but I’m not sure

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What is the recursive formula for 3,7,11,15,...
scoundrel [369]

Arithmetic sequence: a_n=a_1+(n-1)d where: d = common difference between terms; n = term; a_1 = first term of sequence a_n = 3+(n-1)4 = 3+(4n-4) = 4n-1 Therefore, a_50 = 4(50)-1 = 200 - 1 = 199



8 0
4 years ago
The 99 percent confidence interval for the average weight of a product is from 71.36 lbs. to 78.64 lbs. Can we conclude that μ i
Ilya [14]

Answer:

Step-by-step explanation:

Hello!

To decide over a hypothesis test using a confidence interval the following conditions are to be met:

1. The hypotheses should be two-tailed.

2. The hypotheses and the confidence interval should be made for the same parameter.

3. The confidence level and the significance level should be complementary, so if the CI is 1-α: 0.99 the level of significance should be α: 0.01

If all conditions are met, the decision rule is the following:

If the value stated in the null hypothesis is contained by the CI, the decision is to not reject the null hypothesis.

If the value stated in the null hypothesis is not contained by the CI, the decision is to reject the null hypothesis.

Let's say in this example the hypotheses are:

H₀: μ = 71

H₁: μ ≠ 71

α: 0.01

99% Confidence level is (71.36;78.64).

As you can see, 71 is not contained in the CI so the decision to take is to reject the null hypothesis, in other words, the population mean is not 71.

I hope it helps!

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3 years ago
Solve using substitution.<br> y = -x - 1<br> y = x + 3<br> Submit
Alex Ar [27]

Answer:

Step-by-step explanation:

-x - 1 = x + 3

-2x - 1 = 3

-2x = 4

x = -2

y = -2 + 3

y = 1

(-2, 1)

4 0
3 years ago
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