What you do is divide 19.5 by 3 which is 6.5. So every bag would weight 6.5 lbs
Answer:
Tn = 6.4 + 1.8n
Step-by-step explanation:
Given
Sequence: 8.2, 10, 11.8, 13.6
Required
The formula of the sequence.
First, the type of the sequence needs to be determined (arithmetic or geometric)
It is an arithmetic sequence because each successive sequence is separated by a common difference..
The common difference is represented by d and it's calculated as follows.
d = 10 - 8.2 or 11.8 - 10 or 13.6 - 11.8
Each of the above gives
d = 1.8
Now, that we have the common difference; the next is to determine the formula using the Arithmetic Progression formula.
Tn = T1 + (n - 1)d
Where T1 is the first term of the progression; T1 = 8.2
By substituting 8.2 for T1 and 1.8 for d.
This gives
Tn = 8.2 + (n - 1) * 1.8
Open bracket
Tn = 8.2 + 1.8 * n - 1 * 1.8
Tn = 8.2 + 1.8n - 1.8
Collect like terms
Tn = 8.2 - 1.8 + 1.8n
Tn = 6.4 + 1.8n
Hence, the formula of the sequence is Tn = 6.4 + 1.8n
9514 1404 393
Answer:
C. 3x∛(y²z)
Step-by-step explanation:
The relevant rules of exponents are ...
![\sqrt[n]{a^m} = a^\frac{m}{n}\\\\(a^b)^c=a^{bc}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%20%3D%20a%5E%5Cfrac%7Bm%7D%7Bn%7D%5C%5C%5C%5C%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D)
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Your expression can be rewritten and simplified as follows.
![\displaystyle\sqrt[3]{27x^3y^2z}=(3^3x^3y^2z)^\frac{1}{3}=3xy^\frac{2}{3}z^\frac{1}{3}=3x(y^2z)^\frac{1}{3}=\boxed{3x\sqrt[3]{y^2z}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csqrt%5B3%5D%7B27x%5E3y%5E2z%7D%3D%283%5E3x%5E3y%5E2z%29%5E%5Cfrac%7B1%7D%7B3%7D%3D3xy%5E%5Cfrac%7B2%7D%7B3%7Dz%5E%5Cfrac%7B1%7D%7B3%7D%3D3x%28y%5E2z%29%5E%5Cfrac%7B1%7D%7B3%7D%3D%5Cboxed%7B3x%5Csqrt%5B3%5D%7By%5E2z%7D%7D)
A = 1/2bh...multiply both sides by 2
2A = bh...divide both sides by b
(2A)/b = h
s = 1/2gt^2...multiply both sides by 2
2s = gt^2...divide both sides by t^2
(2s) / (t^2) = g
H = vt - 5t^2...add 5t^2 to both sides
H + 5t^2 = vt...divide both sides by t
(H + 5t^2) / t = v
