Answer:
The answer to your question is:
x = 4
y = -1
z = -3
Step-by-step explanation:
3 x + 2 y + z = 7
5 x + 5 y + 4 z = 3
3 x + 2 y + 3 z = 1
![\left[\begin{array}{ccc}3&2&1\\5&5&4\\3&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%261%5C%5C5%265%264%5C%5C3%262%263%5Cend%7Barray%7D%5Cright%5D)
= 45 + 10 + 24 - (30 + 24 + 15)
= 79 - 69
Δ = 10
![\left[\begin{array}{ccc}7&2&1\\3&5&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%262%261%5C%5C3%265%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D)
= 105 + 6 + 8 - (18 + 56 + 5)
= 119 - 79
Δx = 40
![\left[\begin{array}{ccc}3&7&1\\5&3&4\\3&1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%267%261%5C%5C5%263%264%5C%5C3%261%263%5Cend%7Barray%7D%5Cright%5D)
= 27 + 5 + 84 - ( 105 + 12 + 9)
= 116 - 126
Δy = -10
![\left[\begin{array}{ccc}3&2&7\\5&5&3\\3&2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%267%5C%5C5%265%263%5C%5C3%262%261%5Cend%7Barray%7D%5Cright%5D)
= 15 + 70 + 18 - (10 + 18 + 105)
= 103 - 133
= -30
Δz = -30
x = Δx /Δ = 40/10 = 4
y = Δy/Δ = -10/10 = -1
z = Δz/Δ = -30/10 = -3
Are there any other numbers to this question? I will be glad to help.
Answer:
very less
Step-by-step explanation:
When rolling a fair 6 sided die 5 times there are 6^5 possible combinations.
Of those, there are 5 ways to roll exactly four sixes, and each of those ways have 5 possible results on the die which has not rolled a six, plus there is 1 way to roll five sixes. 5*5+1=26 combinations with at least four sixes.
Now, take the number of results which fit your conditions, divide by the number of possible results and you get your answer: 26/7776 which can be reduced to 13/3338, a bit less than 0.4%
Who says V1=V2?
if we simplify we get
(2/3)pir₁³=12pir₂²
for V1 to equal V2
a.
solve for r₁ to find r₁ as a function of r₂
(2/3)pir₁³=12pir₂²
times 3/2 both sides and divide by pi
r₁³=18r₂²
cube root both sides
r₁=∛(18r₂²)
if solve for r₂
(2/3)pir₁³=12pir₂²
divide by 12pi both sides
(1/18)r₁³=r₂²
squer root both sides
√((1/18)r₁³)=r₂
double radius of pond which is r1
√((1/18)r₁³)=r₂
r₁ turns to 2r₁ to double radius
√((1/18)(2r₁)³)=r₂double
√(8(1/18)(r₁)³)=r₂double
(√8)(√((1/18)(r₁)³))=r₂double
√((1/18)r₁³)=r₂ so
(√8)(r₂)=r₂double
(2√2)(r₂)=r₂double
the radius of the tank is multipled by 2√2