The Area of the square is (side)^2
side is given to be 2^(4/9)
SO,
side^2 = [2^(4/9)]^2 = 2^(4/9 * 2) = 2^(8/9)
so the area of square is 2^(8/9) sq. inches.
Answer:
![\sqrt[]{\frac{x+8}{4}}-3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3)
Step-by-step explanation:
First rewrite 
as y
Now swap y and x
Add 8 on both sides.
Divide by 4.
Extract the square root on both sides.
![\sqrt[]{\frac{x+8}{4}}=\sqrt[]{(y+3)^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D%3D%5Csqrt%5B%5D%7B%28y%2B3%29%5E2%7D)
![\sqrt[]{\frac{x+8}{4}}=y+3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D%3Dy%2B3)
Subtract 3 on both sides.
![\sqrt[]{\frac{x+8}{4}}-3=y+3-3](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3%3Dy%2B3-3)
![\sqrt[]{\frac{x+8}{4}}-3=y](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7Bx%2B8%7D%7B4%7D%7D-3%3Dy)
14) The sum of anything and its additive inverse is zero. (That is the definition of additive inverse.) The appropriate choice is ...
... A) equal to 0
Answer:
I'm newbie here
Step-by-step explanation:
Elimination Method.
4x + 6/y = 15 → step 1
6x - 8/y = 14 → step 2
so,
4x + 6/y = 15 |×6|
6x - 8/y = 14 |×4|
24x + 6/y(6) = 90
24x - 8/y(4) = 56
24x + 36/y = 90
24x - 32/y = 56
____________ _
68/y = 34
68 = 34y
34y = 68
y = 2
subsitution y = 2 to..
4x + 6/y = 15
4x + 6/2 = 15
4x + 3 = 15
4x = 12
x = 3
So, for x is 3, and for y is 2
The amount of money I spend putting gas in my car this week is
B = 3.079 G
where
' B ' is the number of Bux I spend on gas this week.
' G ' is the number of Gallons I buy this week.
-- The independent variable is 'G'. I can buy as few or as many gallons
as I want to.
-- The dependent variable is ' B '. It <em>depends</em> on how many gallons I buy.
