How many pounds of water must be evaporated from 50 pounds of a 3% salt solution so that the the remaining solution will be 5 %

The maximum number of identical dental care bags the dentist can create with the smallest leftover is 1.

The counts of bags are 2, 4, and 3, respectively.

Clarification:

Total tubes of toothpaste = 56

Total packets of floss = 112

Total toothbrushes = 85

Therefore, the greatest number of identical dental care bags the dentist can assemble with minimal leftover for toothbrushes is found by calculating the H.C.F of 56, 112, 85 = 1 ( ∵ as they are coprime to each other )

Given that there is no common factor among 56, 112, and 85, except for the commonality between 56 and 112, which is 56, the remainder of toothbrushes will be 85-56=29.

Thus considering 28,[TAG_24]]

The numbers can be expressed as follows: 56 = 28×2, 112 = 28×4, and 84 = 28×3

Consequently, the dentist can create 28 bags containing all three supplies, with no leftover from toothpaste and floss, but with leftover from toothbrushes as we substitute 84 for 85 to maintain divisibility by 28 for common factor demonstration.

This results in 2 bags for toothpaste, 4 bags for floss, and 3 bags for toothbrushes.

In response to the earlier inquiry, the diagram depicts horizontal lines along with a vertical segment; FB measures 3. If C is defined as 3, then the coordinates for point D would be (-6,-4). I trust this response has clarified your question.

Answer:

Indeed, the equation is solvable by factoring. By applying the given equation, you can take the square root of both sides. Since both 169 and 9 are perfect squares, the left-hand side simplifies to plus or minus 13/3, producing rational results. Adding 6 to 13/3 yields a rational number while subtracting it does too. Thus, a quadratic equation is factorable if its solutions are rational.