<span>greater than is >.
</span><span>score greater than a 70 on his test to pass
</span><span>
so answer
</span>x > 70
Answer:
1). Number of kids on the other side = 16
2). Brian took the time to each piece of pie = 20 seconds
Step-by-step explanation:
1). Let the number of kids on one side = x
Then the number of kids on other side = 2x
3 adults helped the side having number of kids = x, then the number of people on this side = x + 3
Since the total number of people on this side = 11
Then the equation will be,
x + 3 = 11
x = 11 - 3
x = 8
Then number of kids on the other side = 2x = 2×8
= 16
2). By unitary method,
∵ Brian took the time to eat 6 pieces of the pie = 2 minutes
∴ Brian will take time to eat 1 piece of pie =
minutes
=
seconds
= 20 seconds
32 cm (I hope this helps you)
Answer:
16-30i
Step-by-step explanation:
This is what I got :)
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.