True, it has a 90 degree angle. It’s clear as day. And the Pythagorean theorem works perfectly fine with the two legs and hypotenuse
Since 1 litre is 1000 millilitres, we can convert it to 1000 millilitres. So, we can subtract 375 millilitres from 1000 millilitres. Which will give us the answer of 625 millilitres.
Answer:
26375
On the graph, it tells you how much fuel is left after 2 hours of the journey.
If we take 422,000 and subtract 369,250 from it we get 52,750. However, this is how much gas is used in 2 hours, so to find one hour we need to divide by 2, getting us 26,375.
Therefore, the Jet uses 26,375 gallons per hour during the 16 hour trip from Los Angeles, CA to Melbourne, Australia.
i did the test
Step-by-step explanation:
Answer:
The common difference is 1/16
Step-by-step explanation:
The given sequence is linear. This means that it is an arithmetic sequence. In an arithmetic sequence, each successive term differ by a common difference. The given set of numbers is
1/4, 5/16, 3/8,....
The second term - the first term = the third term - the second term
The common difference would be
5/16 - 1/4 = 3/8 - 5/16 = 1/16
Answer:
IMPOSSIBLE
Step-by-step explanation:
First we set the equation system:
Now we set the matrix in order to have a solution for the system:
![\left[\begin{array}{ccc}1&1&0\\0&4&1\\4a&b&c\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%264%261%5C%5C4a%26b%26c%5Cend%7Barray%7D%5Cright%5D)
Now we are going to apply Gauss-Jordan to find the solution of the system in terms of a, b and c:
![-4aR_{1}+R_{3}\rightarrow R_{3}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4&1\\0&(-4a+b)&c\end{array}\right]](https://tex.z-dn.net/?f=-4aR_%7B1%7D%2BR_%7B3%7D%5Crightarrow%20R_%7B3%7D%5C%5C%5C%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%264%261%5C%5C0%26%28-4a%2Bb%29%26c%5Cend%7Barray%7D%5Cright%5D)
Next step:
![(4a-b)R_{2}+4R_{3} \rightarrow R_{3}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4&1\\0&0&(4a-b+c)\end{array}\right]](https://tex.z-dn.net/?f=%284a-b%29R_%7B2%7D%2B4R_%7B3%7D%20%5Crightarrow%20R_%7B3%7D%5C%5C%5C%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%264%261%5C%5C0%260%26%284a-b%2Bc%29%5Cend%7Barray%7D%5Cright%5D)
Next step:
![(4a-b+c)R_{2}-R_{3} \rightarrow R_{2}\\\\{\left[\begin{array}{ccc}1&1&0\\0&4(4a-b+c)&0\\0&0&(4a-b+c)\end{array}\right]](https://tex.z-dn.net/?f=%284a-b%2Bc%29R_%7B2%7D-R_%7B3%7D%20%5Crightarrow%20R_%7B2%7D%5C%5C%5C%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C0%264%284a-b%2Bc%29%260%5C%5C0%260%26%284a-b%2Bc%29%5Cend%7Barray%7D%5Cright%5D)
Next step:
![4(4a-b+c)R_{1}-R_{2} \rightarrow R_{1}\\\\{\left[\begin{array}{ccc}4(4a-b+c)&0&0\\0&4(4a-b+c)&0\\0&0&(4a-b+c)\end{array}\right]](https://tex.z-dn.net/?f=4%284a-b%2Bc%29R_%7B1%7D-R_%7B2%7D%20%5Crightarrow%20R_%7B1%7D%5C%5C%5C%5C%7B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%284a-b%2Bc%29%260%260%5C%5C0%264%284a-b%2Bc%29%260%5C%5C0%260%26%284a-b%2Bc%29%5Cend%7Barray%7D%5Cright%5D)
With this solution, we have a new equation system:
This system can be solved by Cramer's rule, by finding the matrix determinant:
![\left[\begin{array}{ccc}16&-4&4\\16&-4&4\\4&-1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D16%26-4%264%5C%5C16%26-4%264%5C%5C4%26-1%261%5Cend%7Barray%7D%5Cright%5D)
As the determinant is zero, we can say that the second system is imposible to solve.
