What is the estimated product of two numbers that are not multiples of ten is 2800?

Suppose you buy 1 ticket for $1 out of a lottery of 1000 tickets where the prize for the one winning ticket is to be $1000. wh

qaws [65]

The answer is 1,000,000 because if you multiply 1,000 and 1,000 you will get 1,000,000

8 0

3 years ago

The polygons are similar. The area of one polygon is given. Find the area of the other polygon.

Fantom [35]

Answer:

108 ft²

Step-by-step explanation:

For the polygon on the left, we are given the length of one side (3 ft) and the total area (27 ft²). With this information, we can determine the length of the other side - 9 ft. I found this by dividing 27 ft² by 3 ft, giving me the final result of 9 ft.

Next, we have to find the other side of the polygon on the <em>right</em> so we can ultimately determine its area. It looks like there is a scale factor of 2 between the two polygons, since 3 × 2 = 6. We know that the bottom side of the left polygon is 9, so multiplying 9 by 2 should give us the bottom side of the polygon on the right. 9 × 2 = 18.

Now, we have the side lengths for the polygon on the right and can determine its area. What is 6 ft × 18 ft? Well, the answer is 107 ft², and this is the answer to the question.

Hopefully that's helpful! :)

5 0

2 years ago

Find the number of ways of arranging the numbers

Doss [256]

First of all, note that all integers are either 0,1, or 2 modulo 3 (if you're not familiar with this terminology, it means that every integer is either a multiple of 3, or it is 1 or 2 away from a multiple of 3).

So, we can think of our numbers as

$\begin{array}{c|c}x&x\mod 3\\0&0\\1&1\\2&2\\3&0\\4&1\\5&2\\6&0\\7&1\\8&2\\9&0\end{array}$

In order to make sure that the sum of any three adjacent numbers is divisible by 3, we have to make sure that any group of 3 three adjacent numbers contains a 0, a 1 and a 2. This is possible only if we arrange our 9 numbers in 3 groups of 3 numbers containing 0,1 and 2 exactly once, repeating always the same pattern.

For example, we could arrange our numbers following the pattern

$0,1,2,0,1,2,0,1,2$

or

$2,0,1,2,0,1,2,0,1$

We have $3!=6$ possible patterns. Suppose for example that we choose the pattern

$0,1,2,0,1,2,0,1,2$

One possible way of following this pattern would be the arrangement

$3,1,2,6,4,5,9,7,8$

In fact, we substituted every '0' with a multiple of 3 (3, 6 or 9), every '1' with a number 1 away from a multiple of 3 (1, 4 or 7) and every '2' with a number 2 away from a multiple of 3 (2, 5 or 8).

This means that, once we fix a patter, we have 3 choices for the first 3 slots, 2 choices for the next 3 slots, and the final slot will be fixed. So, we have

$3\cdot 3\cdot 3\cdot 2 \cdot 2 \cdot 2 = 216$