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kipiarov [429]
2 years ago
5

P is the set of positive integer factors of 20, and Q is the set of positive integer factors of 12. If x is a member of set P an

d y is a member of set Q what is the greatest possible value of x-y?
Mathematics
1 answer:
Leokris [45]2 years ago
8 0

Answer:

please check the attachments for better explanation

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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
Sally is buying new school supplies. She used a 15% off coupon on a set of markers that was originally $3.00. She also bought so
fenix001 [56]
The original price of the paper was $3.50

I hope this helps! let me know if you need help with anything else! I'd be happy to help you in anyway I can! :)
6 0
3 years ago
What does f(3) means ?
Butoxors [25]

F(x)=x^2+x+5  

then  

F(3) would be the function evaluated at x=3, or 17.

anytime you see F(3) it is just whatever formula you are looking at evaluated when the variable is equal to 3.

5 0
3 years ago
In kite ABCD, mZBAE = 28º and mZBCE = 58°. Find mZABE.<br> B<br> E<br> А<br> С<br> D<br> mLABE =
Digiron [165]

Answer:

m∠ABE = 62°

Step-by-step explanation:

Since, the given quadrilateral is a kite, diagonals will intersect at 90°.

Therefore, m∠AEB = m∠CEB = 90°

m∠BAE = 28° [Given]

m∠BCE = 58° [Given]

From ΔABE,

m∠BAE + m∠BEA + m∠ABE = 180°

28° + 90° + m∠ABE = 180°

m∠ABE = 180° - 118°

             = 62°

Therefore, measure of angle ABE = 62°.

3 0
3 years ago
what is the coordinate of the vertex of the parabola y-2=1/12(x+10) a. ( 10,-2) b. (-10,2) c. (-2,10) d. (2.-10)
solniwko [45]

Answer:

y-a= c(x-b)^2

With the vertex V= (b,a)

We see that b = -10 and a = 2 and then the vertes wuld be:

V= (-10,2)

And the best option is:

b. (-10,2)

Step-by-step explanation:

For this problem we have the following function:

y-2 = \frac{1}{2} (x+10)

And if we compare this expression with the general expression for a parabola given by:

y-a= c(x-b)^2

With the vertex V= (b,a)

We see that b = -10 and a = 2 and then the vertes wuld be:

V= (-10,2)

And the best option is:

b. (-10,2)

4 0
3 years ago
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