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Answer:

[tex]G' = (1,4)[/tex]

[tex]H' = (-4,5)[/tex]

[tex]I = (-8,2)[/tex]

Step-by-step explanation:

Given

[tex]G = (4,1)[/tex]

[tex]H = (5,-4)[/tex]

[tex]I = (2,-8)[/tex]

Reflection: [tex]y = x[/tex]

Required

Determine the coordinates of G'H'I'

The following applies when a line A is reflected over [tex]y = x[/tex]

[tex]A = (x,y)[/tex]

[tex]A' = (y,x)[/tex]

i.e, we simply swap the positions of x and y

So, for:

[tex]G = (4,1)[/tex]

[tex]H = (5,-4)[/tex]

[tex]I = (2,-8)[/tex]

The reflections are:

[tex]G' = (1,4)[/tex]

[tex]H' = (-4,5)[/tex]

[tex]I = (-8,2)[/tex]

See attachment

Answer:

He would need to walk the dog 6 more times because 8 times 11 is 88 and if he has already walked the dog 5 times you would subtract that from 11 to get 6. So the answer would be 6

Step-by-step explanation:

8·11= 88

11-5= 6

Answer:  [tex]\frac{8x^3}{27y^6}[/tex]

This is the fraction 8x^3 all over 27y^6

On a keyboard, we can write it as (8x^3)/(27y^6)

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Explanation:

The exponent tells you how many copies of the base to multiply with itself.

We'll have three copies of [tex]\left(\frac{2x}{3y^2}\right)[/tex] multiplied with itself due to the cube exponent on the outside.

So,

[tex]\left(\frac{2x}{3y^2}\right)^3 = \left(\frac{2x}{3y^2}\right)*\left(\frac{2x}{3y^2}\right)*\left(\frac{2x}{3y^2}\right)\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{2x*2x*2x}{(3y^2)*(3y^2)*(3y^2)}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{(2*2*2)*(x*x*x)}{(3*3*3)*(y^2*y^2*y^2)}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{8x^3}{9y^6}\\\\[/tex]

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Or another approach you could take is to cube each component of the fraction. The rule I'm referring to is [tex]\left(\frac{a}{b}\right)^c = \frac{a^c}{b^c}[/tex]

Applying that rule will lead to:

[tex]\left(\frac{2x}{3y^2}\right)^3 = \frac{(2x)^3}{(3y^2)^3}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{2^3*x^3}{3^3*(y^2)^3}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{8x^3}{27y^{2*3}}\\\\\left(\frac{2x}{3y^2}\right)^3 = \frac{8x^3}{27y^6}\\\\[/tex]

Either way you should get 8x^3 all over 27y^6 as one fraction.