Answer: the LCM is 96
Step-by-step explanation: looked it up
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:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0