The answer for that problem is X<26
Let the width =x
length = 6x
p=14x
28=14x
x=2
Answer: The set does not have a solution
Step-by-step explanation:
Adding Equations 1 & 3 we get 5x = 7. This gives x = 7/5
Putting this value of x in eq. 2 we get
-2y + 2z = -1-(7/5) or
2y - 2z = 12/5 or 5y - 5z = 6
Multiplying eq. 1 by 2 we get
4x + 2y - 2z = 6
adding this with eq. 2 we get 5x = 5 or x = 1
As the common solution for x from equations 1&3 does not satisfy eq. 1&2 it comes out that the three equations do not have a common solution.
Same can be verified by using different sets of two equations also.
Answer:
16Km due east of school P
Step-by-step explanation:
Given
A school P is 16km due west of a school Q
Thus, we can say that distance PQ = 16 km.
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now we have to find the bearing of Q from P
As distance is same
distance PQ = distance QP
Thus,
Distance will remain same of 16 km.
For direction,
If Q is west of P, then P will be east of Q
as shown in figure P is west of Q,
now from point P , Q is west P.
Thus,
Bearing of School Q from P is 16Km due east of school P