An ant stands at one vertex of a rectangular sheet of paper. The ant then proceeds to walk all the way around the perimeter of the paper. The dimensions of the sheet of paper are 358 inches by 278 inches. How many inches does the ant travel? Show your work.
The perimeter of a rectangle is equal to the sum of the four sides. In this case, the dimensions of the rectangle are 358 inches by 278 inches, so the ant has to travel a distance of $358\text{ in.} + 278\text{ in.} + 358\text{ in.} + 278\text{ in.} = \boxed{1272\text{ in.}}$.
Determine the sign of the product of −29×17. Enter + for positive or – for negative.
The product of −29×17 is negative. Therefore, the sign of the product is - (negative).
Determine the sign of the product of 1732×(−110)
The product of 1732×(−110) is negative. Therefore, the sign of the product is - (negative).
etermine the sign of the product of −0.708×−6.24.
The product of −0.708×−6.24 is positive. Therefore, the sign of the product is + (positive).
Multiply −5×(−4).
−5×(−4) is equal to 20.
Which of the following shows the Commutative Property of Multiplication?
The Commutative Property of Multiplication states that the order of factors in a multiplication problem can be rearranged without changing the product.
The following equation demonstrates the Commutative Property of Multiplication:
$a \times b = b \times a$
For example, $2 \times 3 = 3 \times 2$
Therefore, the equation that shows the Commutative Property of Multiplication is $a \times b = b \times a$.
Suppose that each dimension of the sheet of paper described in question 1 is increased by one inch. How many inches will the ant travel in this case? Explain your thinking
If each dimension of the sheet of paper is increased by one inch, the new dimensions would be 359 inches by 279 inches.
To find the perimeter of the new paper, we can use the same formula as before: perimeter = (width + length) + (width + length).
Plugging in the new dimensions, we get: perimeter = (359 inches + 279 inches) + (359 inches + 279 inches).
Simplifying, we find that the ant will travel 359 inches + 279 inches + 359 inches + 279 inches = 1276 inches in this case.
Therefore, the ant will travel a total of 1276 inches if each dimension of the paper is increased by one inch.
a second ant walks all the way around a triangle drawn in the dirt. two of the dimension of the triangle are 5 1/10 centimeters and 7 1/5 centimeters. the total distance traveled by the ant is 20 centimeters. what is the length of the third side of the triangle
Let's denote the length of the third side of the triangle as $x$ centimeters.
The total distance traveled by the second ant is the sum of the lengths of all three sides of the triangle: $5\frac{1}{10} + 7\frac{1}{5} + x$.
According to the problem, the total distance traveled by the ant is 20 centimeters, so we can set up the equation:
$5\frac{1}{10} + 7\frac{1}{5} + x = 20$
To solve for $x$, we need to convert the mixed numbers to improper fractions:
$5\frac{1}{10} = \frac{51}{10}$ and $7\frac{1}{5} = \frac{36}{5}$.
Substituting these values into the equation, we have:
$\frac{51}{10} + \frac{36}{5} + x = 20$
Combining like terms, we get:
$\frac{51 + 72}{10} + x = 20$
Simplifying further:
$\frac{123}{10} + x = 20$
To isolate $x$, we subtract $\frac{123}{10}$ from both sides:
$x = 20 - \frac{123}{10}$
To subtract these mixed numbers, we need a common denominator of 10:
$x = \frac{200}{10} - \frac{123}{10}$
$x = \frac{77}{10}$
Therefore, the length of the third side of the triangle is $\boxed{7\frac{7}{10}}$ centimeters.
a third
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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.
Let's assume that the dimensions of the rectangular sheet of paper are $x$ centimeters by $y$ centimeters.
We know that the ant travels around the perimeter of the paper, which is equal to $2x + 2y$ centimeters.
According to the problem, the ant travels between 18 and 19 centimeters. Therefore, we can set up the following inequality:
$18 < 2x + 2y < 19$
Dividing the entire inequality by 2, we get:
$9 < x + y < \frac{19}{2}$
Since the tenths and hundredths digits of the dimensions are non-zero, let's assume $x = 9.05$ centimeters and $y = 0.95$ centimeters.
Plugging in these values, we have:
$9.05 + 0.95 < 9.05 + 0.95 < \frac{19}{2}$
$10 < 10 < \frac{19}{2}$
So, this is one possibility for the dimensions of the sheet of paper: $9.05$ centimeters by $0.95$ centimeters.