Let the length and width of the rectangular sheet of paper be $l$ and $w$ centimeters, respectively. The perimeter of the paper is $2(l+w)$. Since the ant walks around the perimeter exactly three times, the distance it travels is $3 \cdot 2(l+w) = 6(l+w)$ centimeters.
Since the distance is between 18 and 19 centimeters, we have $18 \le 6(l+w) \le 19$. Dividing by 6, we get $3 \le l+w \le \frac{19}{6}$.
Now we need to find a pair of positive decimal numbers $l$ and $w$ that satisfy $3 \le l+w \le \frac{19}{6}$. Let's try taking $l = 1.5$ and $w = 1.5$. Then $l+w = 3$, which satisfies the inequality. Therefore, one possibility for the dimensions of the sheet of paper is $\boxed{1.5 \times 1.5}$.
A third ant walks around the perimeter of another rectangular sheet of paper. The dimensions of this sheet of paper are given in decimal numbers expressed to the hundredth place, with the tenths and hundredths digits being non-zero. If the ant travels between 18 and 19 centimeters, what are the dimensions of the sheet of paper? Give one possibility.
5 answers
what??
I apologize for my previous incorrect response. Let's approach the problem again.
Let the length and width of the rectangular sheet of paper be $l$ and $w$ centimeters, respectively. The ant walks around the perimeter once, which is $2(l+w)$ centimeters. Since the ant walks around the perimeter exactly three times, the distance it travels is $3 \cdot 2(l+w) = 6(l+w)$ centimeters.
The problem states that the distance traveled by the ant is between 18 and 19 centimeters. Therefore, we have $18 \le 6(l+w) \le 19$. Dividing by 6, we get $3 \le l+w \le \frac{19}{6}$.
We want to find positive decimal numbers $l$ and $w$ such that $l+w$ is between 3 and $\frac{19}{6}$. One possible solution is when $l = 1.4$ and $w = 1.6$. In this case, $l+w = 1.4 + 1.6 = 3$, which is between 3 and $\frac{19}{6}$.
Therefore, one possibility for the dimensions of the sheet of paper is $\boxed{1.4 \times 1.6}$.
Let the length and width of the rectangular sheet of paper be $l$ and $w$ centimeters, respectively. The ant walks around the perimeter once, which is $2(l+w)$ centimeters. Since the ant walks around the perimeter exactly three times, the distance it travels is $3 \cdot 2(l+w) = 6(l+w)$ centimeters.
The problem states that the distance traveled by the ant is between 18 and 19 centimeters. Therefore, we have $18 \le 6(l+w) \le 19$. Dividing by 6, we get $3 \le l+w \le \frac{19}{6}$.
We want to find positive decimal numbers $l$ and $w$ such that $l+w$ is between 3 and $\frac{19}{6}$. One possible solution is when $l = 1.4$ and $w = 1.6$. In this case, $l+w = 1.4 + 1.6 = 3$, which is between 3 and $\frac{19}{6}$.
Therefore, one possibility for the dimensions of the sheet of paper is $\boxed{1.4 \times 1.6}$.
can you show the answer without the $ signs
Certainly! One possibility for the dimensions of the sheet of paper is 1.4 centimeters by 1.6 centimeters.
7 answers