The 3rd slot is the answer you are looking for
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.
Step-by-step explanation:
log (√1000000x)
Rewrite √1000000x as (1000000x)1/2.
expand long ((1000000x)1/2) by moving 1/2
oby moving logarithm.
1/2 longth (1000000x)
Rewrite
log
(1000000x) as log(1000000)+log(x).
1/2(log(1000000)+log(x))
Logarithm base 10 of 1000000 is 6.
1/2(6+log(x))
Apply the distributive property.
1/2.6+1/2 log(x)
Cancel the common factor of 2.
3+1/2 long(x)
Combine 1/2 and log(x)
3+ long(x)/2
Answer:
P = 328 yards
Step-by-step explanation:
formula is P= 2l + 2w

Answer:
6. 6x-21=33+9x 7. -15x=-5(3x+7) 8. 16x-19=113-6x 9. -19x-34=56-x 10. -6(4x+3)=6(4x-3)?
6. 6x-21 = 33 +9x
+21 + 21
6x = 54 + 9x
-9x -9x
-3x = 54
/3 /3
x = -18
7. -15x = -5(3x+7)
First multiply the -5 to the other numbers.
-15x = -15x - 35
0x = -35
7 is impossible I believe.
8. 16x-19 = 113-6x
+6 +6
22x - 19 = 113
+ 19 +19
22x = 132
/22 /22
x = 6
9. -19x - 34 = 56 - x
+19x +19x
-34 = 56 + 18x
-56 -56
-90 = 18x
/18 /18
x = -5
10. -6(4x+3) = 6[4x-3]
-24x - 18 = 24x - 18
+18 +18
-24x= 24x
x = 0, since only zero can make the equation true.