To solve this problem you must apply the proccedure shown below:
1. You have that the ellipse given as a vertical major axis (a=13), therefore, taking the ellipse with its center at the origin, you have the following equation:
(y^2/a^2)+(x^2/b^2)=1
2. You have the distance from the center of the ellipse to the focus:
c=12, therefore, you can calculate the value of b, the minor radius:
c^2=a^2-b^2
b=√(13^3-12^2)
b=5
3. Therefore, the equation is:
a^2=169
b^2=25
(y^2/169)+(x^2/25)=1
The answer is: (y^2/169)+(x^2/25)=1
The correct answer is maybe 3
<h3>☂︎ Answer :- </h3>
<h3>☂︎ Solution :- </h3>
- LCM of 5 , 18 , 25 and 27 = 2 × 3³ × 5²
- 2 and 3 have odd powers . To get a perfect square, we need to make the powers of 2 and 3 even . The powers of 5 is already even .
In other words , the LCM of 5 , 18 , 25 and 27 can be made a perfect square if it is multiplied by 2 × 3 .
The least perfect square greater that the LCM ,
☞︎︎︎ 2 × 3³ × 5² × 2 × 3
☞︎︎︎ 2² × 3⁴ × 5²
☞︎︎︎ 4 × 81 × 85
☞︎︎︎ 100 × 81
☞︎︎︎ 8100
8100 is the least perfect square which is exactly divisible by each of the numbers 5 , 18 , 25 , 27 .
X=7
100-30=70
70/10
x=7
hope this helps
