Answer: 31,948.80
Step-by-step explanation: I thin it his is correct. I multiplied the original amount by .04 then subtracted that from the original number.
, and 
. Subtracting this from the numerator gives a remainder of
, and 
. Subtracting this from the previous remainder gives a new remainder of
is not a multiple of 
, so we're done. Then
Answer:
Final answer is ![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3Dx)
.
Step-by-step explanation:
Given problem is ![\sqrt[3]{x}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
.
Now we need to simplify this problem.
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D)
Apply formula
![\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Bx%5Ep%7D%5Ccdot%5Csqrt%5Bn%5D%7Bx%5Eq%7D%3D%5Csqrt%5Bn%5D%7Bx%5E%7Bp%2Bq%7D%7D)
so we get:
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B1%2B2%7D%7D)
![\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%5E1%7D%5Ccdot%5Csqrt%5B3%5D%7Bx%5E2%7D%3D%5Csqrt%5B3%5D%7Bx%5E%7B3%7D%7D)
Hence final answer is .
Answer: The detergent for <u><em>$5.50 for 96 oz</em></u> is better.
Step-by-step explanation:
128 ÷ $6.90 ≈ 18.55
96 ÷ $5.50 ≈ 17.45
1/9
Here's the proper explanation:
Since the vertex is (2,-1), the original equation will look like this: y=a*(x-2)^2-1
When the substitution happens we get to this:
0=a*(5-2)^2-1 then 1=a*(5-2)^2 -> 1=a*9
So, a=1/9! Written in another form:
1/9x^2-4/9x+4/9-1
Hope could help :)
