Answer: the height in inches, of the pile after 3 weeks is 34 11/12 inches
Step-by-step explanation:
Each consecutive week for the next 5 weeks the height of pile increase by 8 7/12 inches. Converting 8 7/12 inches to improper fraction, it becomes 103/12 inches. The height is increasing in an arithmetic progression. The formula for determining the nth term of an arithmetic sequence is expressed as
Tn = a + (n - 1)d
Where
a represents the first term of the sequence.
d represents the common difference.
n represents the number of terms in the sequence.
From the information given,
a = 17 3/4= 71/4 inches
d = 103/12 inches
n = 3 weeks
the height in inches, of the pile after 3 weeks, T3. Therefore,
T3 = 71/4 + (3 - 1)103/12
T3 = 71/4 + 2 × 103/12 = 71/4 + 103/6
T3 = 419/12 inches = 34 11/12 inches
From the right hand side, we will need to find a way to rewriting 3x²y in terms of cube roots.
We know that 27 is 3³, so if we were to rewrite it in terms of cube roots, we will need to multiply everything by itself two more twice. (ie we can rewrite it as ∛(3x²y)³)
Hence, we can say that it's:
![\sqrt[3]{162x^{c}y^{5}} = \sqrt[3]{(3x^{2}y)^{3}} * \sqrt[3]{6y^{d}}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B162x%5E%7Bc%7Dy%5E%7B5%7D%7D%20%3D%20%5Csqrt%5B3%5D%7B%283x%5E%7B2%7Dy%29%5E%7B3%7D%7D%20%2A%20%5Csqrt%5B3%5D%7B6y%5E%7Bd%7D%7D)
![= \sqrt[3]{162x^{6}y^{3+d}}](https://tex.z-dn.net/?f=%3D%20%5Csqrt%5B3%5D%7B162x%5E%7B6%7Dy%5E%7B3%2Bd%7D%7D)
Hence, c = 6 and d = 2
Let A=(0,0)(x₁,x₂), B=(6,0)(x₂,y₂) and C=(0,6)(x₃,y₃)
Centroid of ΔABC is given by,
G(x,y) = [x₁+x₂+x₃/3 , y₁+y₂+y₃/3] = [0+6+0/3 , 0+0+6/3] = [2,2]