It would be 67,437,000,000 since the eight in 67,436,(8)28,104 rounds the 6 up to a 7.
D. X = 10
6x-20= 40
6x=40+20
6x=60
X=10
Answer:
The weight of container C is 2.1kg.
Step-by-step explanation:
This question is solved using a system of equations.
I am going to say that:
x is the weight of container A.
y is the weight of container B.
z is the weight of container C.
The average weight of 3 containers A,B and C is 3.2kg.
This means that the total weight is 3*3.2 = 9.6kg. So
![x + y + z = 9.6](https://tex.z-dn.net/?f=x%20%2B%20y%20%2B%20z%20%3D%209.6)
Container A is twice as heavy as container B.
This means that
.
Containers B is 400 g heavier than container C.
400g is 0.4kg. So
This means that
, or ![z = y - 0.4](https://tex.z-dn.net/?f=z%20%3D%20y%20-%200.4)
Replacing y and z as functions of x in the first equation:
![x + y + z = 9.6](https://tex.z-dn.net/?f=x%20%2B%20y%20%2B%20z%20%3D%209.6)
![2y + y + y - 0.4 = 9.6](https://tex.z-dn.net/?f=2y%20%2B%20y%20%2B%20y%20-%200.4%20%3D%209.6)
![4y = 10](https://tex.z-dn.net/?f=4y%20%3D%2010)
![y = \frac{10}{4} = 2.5](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B10%7D%7B4%7D%20%3D%202.5)
Container C
![z = y - 0.4 = 2.5 - 0.4 = 2.1](https://tex.z-dn.net/?f=z%20%3D%20y%20-%200.4%20%3D%202.5%20-%200.4%20%3D%202.1)
The weight of container C is 2.1kg.
You can use the keep it, change it, and flip it method. Keep the 8/9. Change the division to multiplication. Flip -14/15 (it becomes -15/14). So, your answer would become 20/21.
The solution is (8,8) which means that at 8 miles both cabs will cost $8