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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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A) Two angles are adjacent if their measures add up to 90°.
slavikrds [6]

Answer:

A) False

B) False

Step-by-step explanation:

A. Two angles whose measures add up to 90° are said to be complementary, while two angles are adjacent when formed beside each other. So, adjacent angles might be complementary or supplementary. Thus, the answer to the first part of the question is false.

B. When two angles share a side, they are said to be adjacent to each other. But supplementary angles add up to 180^{o}. Since adjacent angles might be complementary or supplementary, then the answer to the second part of the question is false.

4 0
3 years ago
HELP I NEED HELPP! Evaluate the expression for the given values.
elena-s [515]

Answer:

When x = 12, y = 4, and z = 2, 8x + 2yz = 112.

Step-by-step explanation:

8x + 2yz

Substitute variables given:

x = 12

y = 4

z = 2

8(12) + 2(4)(2)

Multiply 8 and 12.

96 + 2(4)(2)

Multiply 2 and 4.

96 + 8(2)

Multiply 8 and 2.

96 + 16

Add 96 and 16.

112.

8 0
3 years ago
Can anyone solve these? Pre-Calc BC
viva [34]

Answer:

79. [7, 12], [1, 6]

80. The total revenue in the month of May is $11,575

81. The total revenue in the month of November is $4,630

82. Values obtained from the model and actual values are almost the same, so the given model is precise.

Step-by-step explanation:

79.

The domain of a function is the set of x-values that we are allowed to plug into our function.

The graph below represents f(x),

the graph that the domain of the first part of the function is [7,12]. The domain of the second part of the function is [1,6]. The solution is:

f(x)=\left \{ {{-1.97x+26.3, 7 \leq x \leq 12} \atop {0.505x^{2} -1.47x+6.3, 1 \leq x \leq 6}} \right. \\\\

80.

1 ≤ 5 ≤ 6, so we'll plug the value 5 in the function f(x) = 0.505x² - 1.47x + 6.3

f(x) = y = (0.505)(5)² - 1.47(5) + 6.3

f(x) = y = (0.505)(25) - 7.35 + 6.3

f(x) = y = 12.625 - 1.05

f(x) = y = 11.575

For x=5 which represents 5th month May, we got y=11.575

Therefore, the total revenue in the month of May is 11.575 thousands dollars, ie 11,575 dollars.

81.

7 ≤ 11 ≤ 12, so we'll plug the value 11 in the function f(x) = -1.97x + 26.3

f(x) = − 1.97(11) + 26.3

     = − 21.67 + 26.3

     = 4.630.

For x=11 which represents 11th month November, we got y=4.630 Therefore, the total revenue in the month of November is 4.630 thousands dollars, 4,630 dollars.

82.

Values obtained from the model are y=11.575 (x=5) and y=4.630 (x=11). Actual data values are y=11.5 (x=5) and y=4.4 (x=11)

By comparing these values, we can conclude that the model is precise, because these values are almost the same.

Learn more about Piecewise Functions here: brainly.com/question/13882670

3 0
2 years ago
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