A hyperbola has vertices (±5, 0) and one focus (6, 0). what is the standard-form equation of the hyperbola?
1 answer:
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Since you know a = 5√2, start by plugging that into your other equation, 2a = √2x to get an equation where x is the only variable you need to solve for:
There are two ways of solving for x:
1) Remember when you simplify radicals, you break what's underneath the radical into smaller factors. If a factor repeats under the square root, you can take two of them out and put one of them outside the radical sign (see picture for example). For this problem, you would be doing the reverse of simplifying. You have a 10 on the outside on the left of the equal sign, but a 2 underneath on both sides. If the 10 was taken out, that means there had to be two 10's underneath the radical to begin with. Those two tens would have been multiplied since they are factors, so 10 x 10 = 100. That means x = 100.
2) Solve it algebraically. Square both sides, simplify, then divide by 2:
Answer: x = 100
There are two ways of solving for x:
1) Remember when you simplify radicals, you break what's underneath the radical into smaller factors. If a factor repeats under the square root, you can take two of them out and put one of them outside the radical sign (see picture for example). For this problem, you would be doing the reverse of simplifying. You have a 10 on the outside on the left of the equal sign, but a 2 underneath on both sides. If the 10 was taken out, that means there had to be two 10's underneath the radical to begin with. Those two tens would have been multiplied since they are factors, so 10 x 10 = 100. That means x = 100.
2) Solve it algebraically. Square both sides, simplify, then divide by 2:
Answer: x = 100
Answer:
the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k
Explanation:
Entropy is defined as the measure of possible arrangement of the atoms or molecules. Entropy change for the the freezing process mole of 1 liquid methanol at -97.6 degree celcius and 1 atm will be -18.07 J/mol.
The entropy change can be calculated by the given formula:
Thus, the value of change in entropy for freezing process of 1 mole of liquid methanol is -18.07 J/mol.k
(Hope this helps can I pls have brainlist (crown)☺️)
