I think $6.75 for 15 juice boxes is better because, each juice box is 0.45.
While $5.50 for 11 juice boxes is 0.50.
And if we were to add 0.50 enough times to make 15 juice boxes it would equal 7.50
RIP OFFFFFFF
THE SECOND ONE IS A WAY BETTER OF A DEAL
Answer:
(5,2,2)
Step-by-step explanation:
-3x+4y+2z = -3
2x-4y-z=0
y = 3x-13
Multiply the second equation by 2
2*(2x-4y-z)=0*2
4x -8y -2z =0
Add this to the first equation to eliminate z
-3x+4y+2z = -3
4x -8y -2z =0
-------------------------
x -4y = -3
Take the third equation and substitute it in for y
x - 4(3x-13) = -3
Distribute the 4
x - 12x +52 = -3
Combine like terms
-11x +52 = -3
Subtract 52 from each side
-11x +52-52 = -3-52
-11x = -55
Divide by -11
-11x/-11 = -55/-11
x=5
Now we can solve for y
y =3x-13
y =3*5 -13
y = 15-13
y=2
Now we need to find z
2x-4y-z=0
2(5) -4(2) -z=0
10-8 -z=0
2-z=0
Add z to each side
2-z+z= 0+z
2=z
x=5, y=2, z=2
(5,2,2)
Answer:
![\large\boxed{1.\ f^{-1}(x)=\sqrt[12]{3^x}}\\\\\boxed{2.\ f^{-1}(x)=\sqrt[4]{3^x}}\\\\\ \boxed{3.\ f^{-1}(x)=\sqrt[3]{4^{7-x}}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B1.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B12%5D%7B3%5Ex%7D%7D%5C%5C%5C%5C%5Cboxed%7B2.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B4%5D%7B3%5Ex%7D%7D%5C%5C%5C%5C%5C%20%5Cboxed%7B3.%5C%20f%5E%7B-1%7D%28x%29%3D%5Csqrt%5B3%5D%7B4%5E%7B7-x%7D%7D%7D)
Step-by-step explanation:
![2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------](https://tex.z-dn.net/?f=2.%5C%5Cy%3D%5Clog_3x%5E4%5C%5C%5C%5C%5Ctext%7BExchange%20x%20and%20y.%20Solve%20for%20y%3A%7D%5C%5C%5C%5C%5Clog_3y%5E4%3Dx%5CRightarrow3%5E%7B%5Clog_3y%5E4%7D%3D3%5Ex%5CRightarrow%20y%5E%7B4%7D%3D3%5Ex%5C%5C%5C%5Cy%3D%5Csqrt%5B4%5D%7B3%5Ex%7D%5C%5C-------------------------)
![3.\\y=-\log_4x^3+7\\\\\text{Exchange x and y. Solve for y:}\\\\-\log_4y^3+7=x\qquad\text{subtract 7 from both sides}\\\\-\log_4 y^3=x-7\qquad\text{change the signs}\\\\\log_4y^3=7-x\Rightarrow4^{\log_4y^3}=4^{7-x}\\\\y^3=4^{7-x}\Rightarrow y=\sqrt[3]{4^{7-x}}](https://tex.z-dn.net/?f=3.%5C%5Cy%3D-%5Clog_4x%5E3%2B7%5C%5C%5C%5C%5Ctext%7BExchange%20x%20and%20y.%20Solve%20for%20y%3A%7D%5C%5C%5C%5C-%5Clog_4y%5E3%2B7%3Dx%5Cqquad%5Ctext%7Bsubtract%207%20from%20both%20sides%7D%5C%5C%5C%5C-%5Clog_4%20y%5E3%3Dx-7%5Cqquad%5Ctext%7Bchange%20the%20signs%7D%5C%5C%5C%5C%5Clog_4y%5E3%3D7-x%5CRightarrow4%5E%7B%5Clog_4y%5E3%7D%3D4%5E%7B7-x%7D%5C%5C%5C%5Cy%5E3%3D4%5E%7B7-x%7D%5CRightarrow%20y%3D%5Csqrt%5B3%5D%7B4%5E%7B7-x%7D%7D)
<u><em>A</em></u> would be correct hope I helped.
Answer:
=2a
Step-by-step explanation:
a-2/5a and a+2/5a cancel each other out and your left with a+a aka 2a
