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? HELP PLZZZZ
Let's assume the length of the rectangular sheet of paper is $l$ and the width is $w$.
The ant walks around the perimeter of the sheet of paper, which is equal to $2l + 2w$. We know that the ant travels a distance between 18 and 19 centimeters.
Since the tenths and hundredths digits are non-zero, we can assume that both $l$ and $w$ are greater than 1 centimeter.
To find the dimensions, we need to set up the inequality:
$18 < 2l + 2w < 19$
Dividing by 2, we get:
$9 < l + w < \frac{19}{2}$
Since $l$ and $w$ are greater than 1, we can start by trying the smallest possible values for $l$ and $w$ that satisfy the inequality.
Let's set $l = 2$ and $w = 2$.
$l + w = 2 + 2 = 4$, which is not within the range $9 < l + w < \frac{19}{2}$.
Let's try increasing the width to 3.
$l + w = 2 + 3 = 5$, which is still not within the range.
We can continue trying different values until we find the right dimensions.
After trying different values, we find that $l = 5.6$ centimeters and $w = 3.8$ centimeters satisfies the inequality:
$9 < 5.6 + 3.8 < \frac{19}{2}$
This means that the dimensions of the rectangular sheet of paper are approximately 5.6 cm by 3.8 cm.
whats the answer
The dimensions of the rectangular sheet of paper are approximately 5.6 cm by 3.8 cm.
explain without the dolaar sign
The dimensions of the rectangular sheet of paper are 5.6 centimeters by 3.8 centimeters.
can you explain?
Certainly!
To find the dimensions of the rectangular sheet of paper, we need to set up the inequality:
$18 < 2l + 2w < 19$
Since $l$ and $w$ are the length and width of the paper respectively, the expression $2l + 2w$ gives us the perimeter of the paper.
Dividing by 2, we get:
$9 < l + w < \frac{19}{2}$
This inequality tells us that the sum of the length and width of the paper must be between 9 and $\frac{19}{2}$.
Since the tenths and hundredths digits are non-zero, we can assume that both $l$ and $w$ are greater than 1 centimeter, as lengths and widths less than 1 would not satisfy the given conditions.
To find possible values for $l$ and $w$ that satisfy the inequality, we can start by trying the smallest possible values. Let's set $l = 2$ and $w = 2$.
The sum of the length and width is $l + w = 2 + 2 = 4$. However, this value is not within the range $9 < l + w < \frac{19}{2}$.
We can try increasing the width to 3, so $l = 2$ and $w = 3$.
The sum of the length and width is $l + w = 2 + 3 = 5$. However, this value is still not within the desired range.
We need to continue trying different values until we find the right dimensions.
After trying different values, we find that $l = 5.6$ centimeters and $w = 3.8$ centimeters satisfies the inequality:
The sum of the length and width is $l + w = 5.6 + 3.8 = 9.4$, which is within the range $9 < l + w < \frac{19}{2}$.
Therefore, the dimensions of the rectangular sheet of paper are approximately 5.6 centimeters by 3.8 centimeters.
@__@ I didint understand
Apologies for any confusion caused.
To find the dimensions of the rectangular sheet of paper, we need to set up the inequality:
$18 < 2l + 2w < 19$
This inequality represents the distance traveled by the ant walking around the perimeter of the paper. Since the ant travels a distance between 18 and 19 centimeters, we can assume that $2l + 2w$ falls within this range.
Dividing the inequality by 2, we get:
$9 < l + w < \frac{19}{2}$
This means that the sum of the length and width of the paper must be between 9 and $\frac{19}{2}$.
Now, let's try to find the possible values for $l$ and $w$ that satisfy this inequality. We know that both $l$ and $w$ are greater than 1 centimeter, as the tenths and hundredths digits are non-zero.
Starting with the smallest possible values, let's assume $l = 2$ centimeters and $w = 2$ centimeters. In this case, the sum of the length and width is $l + w = 2 + 2 = 4$ centimeters. However, 4 is not within the range of 9 to $\frac{19}{2}$.
Let's try increasing the value of $w$. Assuming $l = 2$ centimeters and $w = 3$ centimeters, the sum of the length and width becomes $l + w = 2 + 3 = 5$ centimeters. However, 5 is still not within the desired range.
We need to keep trying different values. After trying different combinations, we find that $l = 5.6$ centimeters and $w = 3.8$ centimeters satisfy the inequality:
With $l = 5.6$ centimeters and $w = 3.8$ centimeters, the sum of the length and width becomes $l + w = 5.6 + 3.8 = 9.4$ centimeters, which is within the range $9 < l + w < \frac{19}{2}$.
Therefore, the dimensions of the rectangular sheet of paper are approximately 5.6 centimeters by 3.8 centimeters.
To find the dimensions of the sheet of paper, we can use the fact that the ant walks around the perimeter.
Let's assume the length and width of the paper are represented by variables `l` and `w` respectively.
The perimeter (`P`) of a rectangle is given by the formula:
P = 2(l + w)
In this case, the ant walks between 18 and 19 centimeters. Since the dimensions are given in decimal numbers expressed to the hundredth place, we can write the inequality as:
18 ≤ P ≤ 19
Substituting the formula for the perimeter, we have:
18 ≤ 2(l + w) ≤ 19
Now, let's divide the inequality by 2 to solve for (l + w):
9 ≤ l + w ≤ 9.5
Based on the given information, we know that the tenths and hundredths digits of the dimensions are non-zero. Since both `l` and `w` are positive, the smallest possible value for (l + w) is when both `l` and `w` are equal to 4.5.
So, (l + w) must be at least 9, meaning the sum of the length and width of the rectangle must be at least 9 centimeters.
To find the largest possible value for (l + w), we need to consider the greatest possible values for `l` and `w`. Given that the tenths and hundredths digits cannot be zero, we can assume the largest value for `l` and `w` is 9.9.
So, the largest possible value for (l + w) is 9.9 + 9.9 = 19.8.
Thus, (l + w) must not exceed 9.5.
Combining both inequalities, we have:
9 ≤ l + w ≤ 9.5
Since the dimensions of the paper are in decimal numbers expressed to the hundredth place, both `l` and `w` must be less than 9.5. Therefore, the dimensions of the paper could be any pair of positive decimal numbers whose sum falls between 9 and 9.5.
For example, the dimensions could be 4.5 cm x 4.5 cm, 5 cm x 4 cm, 6 cm x 3 cm, etc. There are countless possibilities, as long as the sum of the length and width falls between 9 and 9.5 centimeters.