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elena-s [515]
3 years ago
5

What is the measure of angle CAE??

Mathematics
1 answer:
Luden [163]3 years ago
4 0
The answer is B (120)
You might be interested in
Which expression is it equivalent to?
horrorfan [7]
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.
8 0
3 years ago
What is the solution of -6x+15=-3
drek231 [11]
The answer to your problem is x=3
7 0
3 years ago
Read 2 more answers
An ice cream shop sold 12,789 chocolate and 9,324 cookie dough cones. They sold 1,078 more peanut butter cones than cookie dough
juin [17]
12,789 chocolate
9,324 Cookie Dough
10,402 Peanut Butter Cones
13,788 Vanilla

46,303 in total

3 0
2 years ago
A jeweler has 36 inches of silver chain. She needs 5 time that much to make some necklaces. How much silver chain does the jewel
Elina [12.6K]
36 *<span> 5 = 180 
hope this helps




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4 0
3 years ago
A moose population is growing exponentially following the pattern in the table shown below. Assuming that the pattern continues,
Alika [10]

The population of the moose after 12 years is 7692

Step-by-step explanation:

The form of the exponential growth function is y=a(b)^{x} , where

  • a is the initial value
  • b is the growth factor

The table:

→  x  :  y

→  0  :  40

→  1   :  62

→  2  :  96

→  3  :  149

→  4  :  231

To find the value of a , b in the equation use the data in the table

∵ At x = 0 , y = 40

- Substitute them in the form of the equation above

∵ 40=a(b)^{0}

- Remember any number to the power of zero is 1 (except 0)

∴ 40 = a(1)

∴ a = 40

- Substitute the value of a in the equation

∴ y=40(b)^{x}

∵ At x = 1 , y = 62

- Substitute them in the form of the equation above

∵ 62=40(b)^{1}

∴ 62 = 40 b

- Divide both sides by 40

∴ 1.55 ≅ b

∴ The growth factor is about 1.55

- Substitute its value in the equation

∴ The equation of the population is y=40(1.55)^{x}

To find the population of moose after 12 years substitute x by 12 in the equation

∵ y=40(1.55)^{12}

∴ y ≅ 7692

The population of the moose after 12 years is 7692

Learn more:

You can learn more about the logarithmic functions in brainly.com/question/11921476

#LearnwithBrainly

7 0
3 years ago
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