Answer:
Step-by-step explanation:
A graphing calculator can help you see the center is (4, 3) and the diameter is 10, so the radius is 5.
_____
Or, you can rearrange the equation to standard form by completing the squares.
x² -8x + y² -6y = 0 . . . . . subtract 6y, group terms
(x² -8x +16) +(y² -6y +9) = 16 +9 . . . . . complete the squares by adding the squares of half the coefficient of the linear term: (-8/2)²=16, (-6/2)²=9.
(x -4)² + (y -3)² = 25 . . . . . rewrite in standard form
Compare this to ...
(x -h)² + (y -k)² = r²
and recognize the center (h, k) is (4, 3), and the radius is √25 = 5.
Answer:
Step-by-step explanation:
pi - 1 = 3.14 - 1 = 2.14
x has to be an integer, and the result must be greater than 1.
1 works.
1 < 2.14*1 < 10 2.14 is between 1 and 10.
2 works
1 < (2.14)*2 < 10
1 < 4.28 < 10
3 works
1 < 2.14 * 3 < 10
1 < 6.42 < 10
4 might work
1 < 2.14*4 < 10
1 < 8.56 < 10
and it does work.
5 can't work
1 < 5 * 2.14 < 10
1 < 10.7 < 10 is not true.
Answer: 1,2,3,4
Answer:
10
Step-by-step explanation:
pi is 3.14...
pi - 1 is 2.14...
question becomes 1<2.14x and 2.14x < 10
1/2.14 < x, 0.46... < x
x < 10/2.14, x < 4.67...
combining
0.46< x < 4.67
x can be 1, 2, 3, 4
Answer: 30 grams
Step-by-step explanation: (50 mL x x%) = (1000 mL x 0.3%)
x = 6%
6% x 500 mL = 30 grams
The calculation for the amount of sodium chloride (NaCl) in a solution depends on the desired concentration. For instance, to prepare 50 ml of a 1M solution, one would need 2.922 g of NaCl. For a 500 ml stock solution for the same, the measurement would be ten times this, or 29.22 g.
The question pertains to preparing a solution of sodium chloride N(aCl). The given information indicates that we have 5.30 mol NaCl L solution. To provide an accurate answer, it's necessary to know the targeted concentration for the 50 ml solution. However, lacking this data, we can consider an example where we want to prepare a 1M solution.
In such a case, using the concept of molarity (mol/L), we'd first establish how many moles of NaCl are needed. For a 1M solution, we'd need 1 mol of NaCl per liter of solution. Therefore, in order to prepare 50 ml (or 0.05 L), we require 0.05 mol of NaCl.
The given information also states that 1 mol of NaCl weighs 58.44 g. Therefore, we would need (0.05 mol) * (58.44 g/mol) = 2.922 g of NaCl for 50 ml of a 1M solution. Therefore, if preparing 500 ml of a stock solution, we would require 10 times this amount, or 29.22 g of NaCl.
This is a general guide as the specific quantity can vary based on the desired concentration of the 50 ml solution. For other concentrations, we would use the same method, simply adjusting the moles of NaCl needed as appropriate.
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Answer:
Holly has 20 quart of potato salad left.
Step-by-step explanation:
Holly has 20 quart of potato salad because she originally has 46 quart of potato but gave her friend 26, subtracting 26 from the original amount of quart of potato she had.