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gtnhenbr [62]
2 years ago
5

What is the following quotient? StartFraction 6 minus 3 (RootIndex 3 StartRoot 6 EndRoot) Over RootIndex 3 StartRoot 9 EndRoot E

ndFraction.
Mathematics
1 answer:
Bezzdna [24]2 years ago
4 0

Exponent properties help us to simplify the powers of expressions.  The quotient of the given expression \dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}} is (2∛3 - ∛18).

<h3>What are the basic exponent properties?</h3>

{a^m} \cdot {a^n} = a^{(m+n)}\\\\\dfrac{a^m}{a^n} = a^{(m-n)}\\\\\sqrt[m]{a^n} = a^{\frac{n}{m}}\\\\(a^m)^n = a^{m\times n}\\\\(m\times n)^a = m^a\times n^a\\\\

Given to us

\dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}}

We will solve the problem using the basic exponential properties,

\dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}}\\\\ = \dfrac{6}{\sqrt[3]{9}} - \dfrac{3(\sqrt[3]{6})}{\sqrt[3]{9}}\\\\ = (6\cdot 3^{-\frac{2}{3}}) - [3 \cdot (2 \cdot 3)^{-\frac{2}{3}}3^{-\frac{2}{3}}]\\\\= (2 \cdot 3 \cdot 3^{-\frac{2}{3}}) - [3 \cdot 2^{-\frac{2}{3}} \cdot 3^{-\frac{2}{3}}3^{-\frac{2}{3}}]\\\\

= [2 \cdot 3^{(1-\frac{2}{3})}] - [2^{\frac{1}{3}}\cdot 3^{(1+\frac{1}{3} - \frac{2}{3})}]\\\\=  [2 \cdot 3^{(\frac{1}{3})}] - [2^{\frac{1}{3}}\cdot 3^{(\frac{2}{3})}]\\\\= 2\sqrt[3]{3} - \sqrt[3]{2}\sqrt[3]{9}\\\\=2\sqrt[3]{3} - \sqrt[3]{18}

Hence, the quotient of the given expression \dfrac{6 - 3(\sqrt[3]{6})}{\sqrt[3]{9}} is (2∛3 - ∛18).

Learn more about Exponents:

brainly.com/question/5497425

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A circle has a diameter with endpoints at (6, 5) and (8, 5). Write the equation for the circle.
mario62 [17]

Answer:

(x-7)^2+(y-5)^2=1

Step-by-step explanation:

The two things that are required to formulate the equation of the circle is the center coordinate and the radius of the circle!

<u>Center of the circle:</u>

  • The center of the circle always lies at the midpoint of the endpoints of its diameter: Let's call the endpoints A(6,5) and B(8,5).

Using the midpoint formula we'll get:

(x_m, y_m) = \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)

(x_m, y_m) = \left(\dfrac{6+8}{2},\dfrac{5+5}{2}\right)

(x_m, y_m) = (7,5)

This is the center coordinate of our circle.

<u>Radius: </u>

The radius of the circle is the distance from the center of the circle to any of the endpoints of the diameter (A or B)

We can use the distance formula:

r = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

r = \sqrt{(x_1-x_m)^2+(y_1-y_m)^2}

r = \sqrt{(6-7)^2+(5-5)^2}

r = \sqrt{1^2}

r = 1

<u>Equation of the circle: </u>

The equation is written as:

(x-a)^2+(y-b)^2=r^2

here, (a,b) are the center points of the circle

in our case this is (a,b)=(x_m,y_m)=(7,5)

and r = 1

(x-7)^2+(y-5)^2=1^2

(x-7)^2+(y-5)^2=1

This is the equation of the circle!

3 0
3 years ago
A senior paid $3.47, $9.50 and $2.50 for lunch during a basketball tournament. What was the average amount he paid over three da
cestrela7 [59]

Answer:

The average amount he paid over 3 days is $15.47.

Step-by-step explanation:

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What is an equation of the line that passes through the points (4,-1) and (8,5)?
fiasKO [112]

Answer:

y=3/2x-7

Step-by-step explanation:

the equation of the line for slope-intercept form is y=mx+b, where  m is the slope and b is the y intercept.

we are given two points: (4,-1) and (8,5)

the equation for slope is (y2-y1)/(x2-x1)

label the points:

x1=4

y1=-1

x2=8

y2=5

now substitute into the equation:

m=(5--1)/(8-4)

m=6/4

m=3/2

the slope of the line is 3/2

here is our equation so far:

y=3/2x+b

we need to find b

since the equation will pass through the points, we can substitute either one into the equation to find b

let's use (4,-1) as an example

substitute into the equation

-1=3/2(4)+b

-1=6+b

-7=b

the y intercept is -7

so the equation is y=3/2x-7

hope this helps!

8 0
3 years ago
Please help. If you could/want please explain how you got it you don’t have to though no pressure
Stolb23 [73]

Answer:

x = 5.14

Step-by-step explanation:

Both angles are equal so

5x-6 = 11x-42

42-6 = 11x - 5x

36 = 7x

x = 5.14

4 0
3 years ago
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