Answer:
The standard parabola
y² = -18 x +27
Length of Latus rectum = 4 a = 18
Step-by-step explanation:
<u><em>Explanation:-</em></u>
Given focus : (-3 ,0) ,directrix : x=6
Let P(x₁ , y₁) be the point on parabola
PM perpendicular to the the directrix L
SP² = PM²
(x₁ +3)²+(y₁-0)² = 
x₁²+6 x₁ +9 + y₁² = x₁²-12 x₁ +36
y₁² = -18 x₁ +36 -9
y₁² = -18 x₁ +27
The standard parabola
y² = -18 x +27
Length of Latus rectum = 4 a = 4 (18/4) = 18
1. Multiply each equation so they end up with the same coefficient
2. Subtract your second equation from the first
3. Solve for one of the variables (I tend to solve an equation that only contains 2 variables if possible. So it would be if you have one question with x and y, solve for the easier one)
4. Substitute the variable you found in the least step into one of the other equations and find a second variable.
5. Substitute both variables you found into the last equation and there you should be left with x, y, and z :))
I hope this helped sksjsk if it didn’t I could write it out to hopefully help more :)
Answer:
Step-by-step explanation:
1 - 8
2 - 14
3 - 22
Complete Question:
Is the value of the fraction 7−2y/6 greater than the value of the fraction 3y−7/12 ? For what values?(Make sure to use an inequality)
Answer:
y < 3
Step-by-step explanation:
The given two fractions are:
and 
We have to tell for which range of values is the value of first fraction larger than the second fraction. This can be done by setting up an inequality as shown:

The range of y which will satisfy this inequality will result in first fraction of larger value as compared to the second fraction.
Multiplying both sides of inequality by 12, we get:

This means, for y lesser than 3, the value of first fraction is larger than the second one.