<span><span><span><span><span><span>x</span><span>=</span><span>2</span><span>(</span><span>y</span><span>+</span><span>3</span><span>)</span></span></span></span></span></span><span>x=<span><span>9</span><span><span>2(y−37)</span></span><span></span>
</span></span><span><span>
</span></span>
Answer:
-4,-3,-2,-1,0,1
Step-by-step explanation:
First, doublepound and simplify it.
-15<3n
3n<6
Solve:
-5<n
n<2
Compound:
-5<n<2.
So the values are -4,-3,-2,-1,0,1
Hope this helps plz hit the crown :D
Answer:
Step-by-step explanation:
Given:
A car starts with a dull tank of gas
1/7 of the gas has been used around the city.
With the rest of the gas in the car, the car can travel to and from Ottawa three times.
Question asked:
What fractions of a tank of gas does each complete trip to Ottawa use?
Solution:
Fuel used around the city =
Remaining fuel after driving around the city = 1 -
=
According to question:
As from the rest of the gas in the car that is , the car can complete 3 trip to Ottawa which means,
By unitary method:
The car can complete 3 trip by using = tank of gas.
The car can complete 1 trip by using =
=
=
= tank of gas
Thus, tank of gas used for each complete trip to Ottawa.
Answer:
11.01
Step-by-step explanation:
3.78+4.51+2.72=11.01