Andre wants to make an open-top box by cutting out corners of a 22 inch by 28 inch piece of poster board and then folding up the

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.

Answer:

(a) What will his age be after 5 years?

"5 + y" years old

(b) What was his age 6 years ago?

"y - 6" years old

(c) If his grandfather’s age is five times his, how old is his grandfather?

"5y" years old

(d) His father is 6 years older than three times his age. How old is his father?

"6 + 3y" years old

Note: Disregard the quotation marks, ""

Answer: F(x) > 0 in the ranges (-0.7, 0.76) and (0.76, ∞).

I hope this is useful:)

Answer:

Given that the frog jumps every 10 seconds

   (using digits from a random number table)

• It requires 7 jumps with 2 in the reverse direction (either left or right) for the frog to get off the board in 60 seconds.
• Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
• Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.

Step-by-step explanation:

A frog positioned right at the center of a 5ft long board is 2.5 ft away from either edge.

Every 10 seconds, the frog jumps left or right.

If the frog's jumps are LLRLRL, it will remain on the board at the leftmost square.

If it jumps as LLRLL, it will jump off the board after fifty seconds.

Given that the frog jumps every 10 seconds

   (using digits from a random number table)

• It requires 7 jumps with 2 in reverse direction (either left or right) for the frog to get off the board in 60 seconds.
• Alternatively, 3 jumps in the same direction will also lead to the frog being off the board.
• Furthermore, it would take 5 jumps with one in the opposite direction within the time limit of 60 seconds to leave the board.