Answer:
Putting the value of x= 2 in the first equation,
6(2)+y= 17
y= 17-12 = 5
This value is same as the value for y given in the question.
Therefore, it satisfies the equation.
Again, putting the value of x= 2 in the second equation,
3(2)+14y= 16
14y= 16-6 = 10
y= 10/14 = 5/7
It doesn't satisfies the 2nd equation
Hence, (2,5) is not the solution to this system.
So one miles for A and the fee is $170 then B is $160 then if you used company A you'd go 3 miles to match company B which their miles would be 2,
which I guess the answer would be 5 miles?
If the are equal then they can be similar, otherwise they are not.
Answer:
f(x)= -0.4x+107
Step-by-step explanation:
Note: The height of the room must be 3 m instead of 3 cm because 3 cm is too small and it cannot be the height of a room.
Given:
Perimeter of the floor of a room = 18 metre
Height of the room = 3 metre
To find:
The area of 4 walls of the room.
Solution:
We know that, the area of 4 walls of the room is the curved surface area of the cuboid room.
The curved surface area of the cuboid is
![C.S.A.=2h(l+b)](https://tex.z-dn.net/?f=C.S.A.%3D2h%28l%2Bb%29)
Where, h is height, l is length and b is breadth.
Perimeter of the rectangular base is 2(l+b). So,
![C.S.A.=\text{Perimeter of the base} \times h](https://tex.z-dn.net/?f=C.S.A.%3D%5Ctext%7BPerimeter%20of%20the%20base%7D%20%5Ctimes%20h)
Putting the given values, we get
![C.S.A.=18\times 3](https://tex.z-dn.net/?f=C.S.A.%3D18%5Ctimes%203)
![C.S.A.=54](https://tex.z-dn.net/?f=C.S.A.%3D54)
Therefore, the area of 4 walls of the room is 54 sq. metres.