Find the coordinates of the x- and y-intercepts for an ellipse with the equation (x+1)^2/9 + (y-2)^2/8 = 1
1 answer:
Answer:
- x-intercepts: -1±√4.5
- y-intercepts: 2±8/3
Step-by-step explanation:
The x-intercepts are where y = 0.
(x +1)^2/9 +(0 -2)^2/8 = 1
(x +1)^2/9 +1/2 = 1 . . . . . simplify a bit
(x +1)^2 = 9/2 . . . . . . . . .subtract 1/2, multiply by 9; next: square root, add -1
x = -1 ±√4.5 . . . . . x-intercepts
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The y-intercepts are where x=0.
(0+1)^2/9 +(y-2)^2/8 = 1
1/9 + (y -2)^2/8 = 1 . . . . . simplify a bit
(y -2)^2 = 64/9 . . . . . . . . subtract 1/9, multiply by 8,
y = 2 ±8/3 . . . . . . take the square root, add 2 . . . . y-intercepts
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Given:
The tens digit of a two digit number is 5 greater the units digit.
If you subtract double the reversed number from it, the result is a fourth of the original number.
To find:
The original number.
Solution:
Let n be the two digit number and x be the unit digit. Then tens digit is (x+5) and the original number is:
Reversed number is:
If you subtract double the reversed number from it, the result is a fourth of the original number.
Multiply both sides by 4.
Divide both sides by 55.
The unit digit is 2. So, the tens digit is .
Therefore, the original number is 72.