Answer:
20 cups
Step-by-step explanation:
Let the small pitchers be given by s
and the larger pitchers by l
According to the given information:
![2s + l = 8....(1) \\ and \\ l - s = 2 \\ l = s + 2...(2) \\ substituting \: l = s + 2 \: in \: equation \: (1) \\ 2s + s + 2 = 8 \\ 3s = 8 - 2 \\ 3s = 6 \\ s = \frac{6}{3} \\ \huge \red{ s = 2} \\ substituting \: s = 2 \: in \: equation \: (2) \\ l = 2 + 2 \\ \huge \purple{ l = 4}\\\\Now\\ 3l +4s \\= 3\times 4+4\times 2\\= 12 +8\\= 20\: cups](https://tex.z-dn.net/?f=2s%20%2B%20l%20%3D%208....%281%29%20%5C%5C%20and%20%5C%5C%20l%20-%20s%20%3D%202%20%5C%5C%20l%20%3D%20s%20%2B%202...%282%29%20%5C%5C%20substituting%20%5C%3A%20l%20%3D%20s%20%2B%202%20%5C%3A%20in%20%5C%3A%20equation%20%5C%3A%20%281%29%20%5C%5C%202s%20%2B%20s%20%2B%202%20%3D%208%20%5C%5C%203s%20%3D%208%20-%202%20%5C%5C%203s%20%3D%206%20%5C%5C%20s%20%3D%20%20%5Cfrac%7B6%7D%7B3%7D%20%20%5C%5C%20%5Chuge%20%5Cred%7B%20s%20%3D%202%7D%20%5C%5C%20substituting%20%5C%3A%20%20%20s%20%20%3D%202%20%5C%3A%20in%20%5C%3A%20equation%20%5C%3A%20%282%29%20%20%5C%5C%20l%20%3D%202%20%2B%202%20%5C%5C%20%5Chuge%20%5Cpurple%7B%20l%20%3D%204%7D%5C%5C%3C%2Fp%3E%3Cp%3E%5C%5CNow%5C%5C%203l%20%2B4s%20%5C%5C%3D%203%5Ctimes%204%2B4%5Ctimes%202%5C%5C%3C%2Fp%3E%3Cp%3E%3D%2012%20%2B8%5C%5C%3C%2Fp%3E%3Cp%3E%3D%2020%5C%3A%20cups)
So, 3 large pitchers and 4 small pitchers hold 20 cups of water.
The numbers from which we are to determine the smallest number that can be divided leaving a remainder of 5 are 44, 55, and 220. We have to do the prime factorization of all the numbers to determine their least common multiple (LCM).
Factorization of 44:
44 = 11 x 4 = 11 x 2 x 2
Factorization of 55:
55 = 11 x 5
Factorization of 220:
220 = 11 x 20 = 11 x 5 x 4
Since 220 has all the factors from 55 and 44 then, the least common multiple is 220. Add 5 to the LCM in order to determine the unknown.
<em>ANSWER: 225</em>
Answer:
The explicit rule of the geometric sequence
aₙ = 187.5 (0.8)ⁿ⁻¹
Step-by-step explanation:
<u><em> Step(i):-</em></u>
Given that the third term of the sequence = 120
tₙ = a rⁿ⁻¹
t₃ = a r³⁻¹ = ar²
120 = ar² ..(i)
Given that the fifth term of the given geometric sequence = 76.8
tₙ = a rⁿ⁻¹
t₅ = a r⁵⁻¹ = a r⁴
76.8 = a r⁴...(ii)
<u><em>Step(ii):</em></u>-
Dividing (ii) and (i)
![\frac{ar^{4} }{ar^{2} } = \frac{76.8}{120}](https://tex.z-dn.net/?f=%5Cfrac%7Bar%5E%7B4%7D%20%7D%7Bar%5E%7B2%7D%20%7D%20%3D%20%5Cfrac%7B76.8%7D%7B120%7D)
r² = 0.64
r =√ 0.64 = 0.8
Substitute r= 0.8 in equation (i)
120 = ar²
120 = a(0.8)²
⇒ ![a = \frac{120}{0.64} =187.5](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B120%7D%7B0.64%7D%20%3D187.5)
<u><em>Step(iii):-</em></u>
The explicit rule of the geometric sequence
aₙ = a rⁿ⁻¹
put a= 187.5 and r = 0.8
aₙ = 187.5 (0.8)ⁿ⁻¹
Answer:
x = -4
Step-by-step explanation:
Solve for x:
3 x - 5 x + 2 = 10
Grouping like terms, 3 x - 5 x + 2 = 2 + (3 x - 5 x):
2 + (3 x - 5 x) = 10
3 x - 5 x = -2 x:
-2 x + 2 = 10
Subtract 2 from both sides:
(2 - 2) - 2 x = 10 - 2
2 - 2 = 0:
-2 x = 10 - 2
10 - 2 = 8:
-2 x = 8
Divide both sides of -2 x = 8 by -2:
(-2 x)/(-2) = 8/(-2)
(-2)/(-2) = 1:
x = 8/(-2)
The gcd of 8 and -2 is 2, so 8/(-2) = (2×4)/(2 (-1)) = 2/2×4/(-1) = 4/(-1):
x = 4/(-1)
Multiply numerator and denominator of 4/(-1) by -1:
Answer: x = -4