Answer:
18
Step-by-step explanation:
A two-digit number (xy) will have the value 10x+y. The sum of its digits is x+y.
You want ...
10x +y = 2(x +y)
1 ≤ x ≤ 9
0 ≤ y ≤ 9
The equation can be simplified to ...
8x = y
The only single-digit values for x and y that satisfy the requirements are ...
x = 1, y = 8
The two-digit number of interest is 18.
_____
Check
Its sum of digits is 1+8 = 9. 2×9 = 18.
_____
Comment on the relationship in the problem
Many of us learned our multiplication tables for 9s by remembering that the sum of digits of 9n was 9, and that the leading digit of the product was n-1.
Answer:
the statement is x=y/8 the number is 18
Step-by-step explanation:
ur welcome.
It cannot be eliminated in a later budget, in order to save money.
b.
Living expenses have a higher priority than any other type of expense.
c.
The spender controls the amount of the rent by choice of apartment.
d.
It is by contract the same amount every month.
Answer:
its D
Step-by-step explanation:
A vertical shift 32 units upward.
A horizontal shift 32 units to the right.
A vertical shift 32 units downward.
A horizontal shift 32 units to the left
Answer:
he will need 7.5 cups of flowers or 7 2/4
Step-by-step explanation:
So basically you have to keep adding 5/4 for every 1/2 you get until you reach 3.So when you do this you get 7.4.
A faster way to do this is to find how 1/2 are in 3 cups and then multiply that by the 5/4.
Answer:
3 3/4
Step-by-step explanation:
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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