Can anyone help me with these problems?

1. A square has a side of 2 decimeters. A new square is formed by joining the midpoints of the sides. If this procedure is repeated 6 times, what is the perimeter and area of the seventh square ?
2. Find the sum of those numbers between 15 and 810 which have 8 as their units digit.
3. In an arithmetic sequence, the sixth term is twice the 3rd term and the 30th term is 60. find the common difference?

#1. Each nested side is 1/√2 as long as the previous side. Use that to find the length of the nth nested side.

#2. Since the last digit is 8, there are 80 of them: 18,28,...808
Now you know the sum of the numbers from 1 to 80 is 80*81/2. So, you have 10 times that sum, plus 80*8, including all the final digits.

#3.
a+5d = 2(a+2d)
a+29d = 60
solve for d.

30

Sure, I can help you with these problems. Let's solve them step-by-step:

1. A square has a side of 2 decimeters. A new square is formed by joining the midpoints of the sides. If this procedure is repeated 6 times, what is the perimeter and area of the seventh square?

To solve this problem, we need to find the pattern of how the side length of the square changes as the procedure is repeated. Each time the procedure is repeated, the new square is formed by joining the midpoints of the previous square.

Let's start with the original square with a side length of 2 decimeters. After the first iteration, the side length of the new square will be the diagonal of the original square divided by 2.

The diagonal of a square with side length 2 decimeters can be found using Pythagoras' theorem:
diagonal = sqrt((side)^2 + (side)^2) = sqrt((2)^2 + (2)^2) = sqrt(8) = 2*sqrt(2) decimeters.

So, the side length of the new square after the first iteration is 2*sqrt(2)/2 = sqrt(2) decimeters.

Now, let's continue this process for 6 iterations:
Iteration 1: Side length = 2 decimeters
Iteration 2: Side length = sqrt(2) decimeters
Iteration 3: Side length = sqrt(2)/2 decimeters
Iteration 4: Side length = sqrt(2)/2 * sqrt(2) = 1 decimeter
Iteration 5: Side length = 1/2 decimeters
Iteration 6: Side length = 1/4 decimeters

For the seventh square, we can see the pattern is that the side length is halved with each iteration. Therefore, for the seventh square, the side length will be 1/8 decimeters.

To find the perimeter of the seventh square, we multiply the side length by 4 (since it's a square):
Perimeter = (1/8) * 4 = 1/2 decimeters.

To find the area of the seventh square, we square the side length:
Area = (1/8)^2 = 1/64 square decimeters.

So, the perimeter of the seventh square is 1/2 decimeters and the area is 1/64 square decimeters.

2. Find the sum of those numbers between 15 and 810 which have 8 as their units digit.

To find the numbers between 15 and 810 with 8 as their units digit, we need to list all the numbers that satisfy this condition. Let's list out these numbers:

18, 28, 38, 48, 58, 68, 78, 88, 98, 108, 118, 128, 138, 148, 158, ..., 808.

To find the sum of these numbers, we can use the formula for the sum of an arithmetic series:
Sum = (number of terms / 2) * (first term + last term).

In this case, the first term is 18, the last term is 808, and there are 8 terms.

Sum = (8 / 2) * (18 + 808) = 4 * 826 = 3304.

So, the sum of the numbers between 15 and 810 with 8 as their units digit is 3304.

3. In an arithmetic sequence, the sixth term is twice the 3rd term and the 30th term is 60. Find the common difference?

To solve this problem, we need to use the formula for the n-th term of an arithmetic sequence:
n-th term = first term + (n - 1) * common difference.

Let's start by using the given information to set up two equations:

Equation 1: 6th term = 2 * 3rd term
Using the n-th term formula, we have:
first term + (6 - 1) * common difference = 2 * (first term + (3 - 1) * common difference)

Simplifying the equation:
first term + 5 * common difference = 2 * (first term + 2 * common difference)

Equation 2: 30th term = 60
Using the n-th term formula, we have:
first term + (30 - 1) * common difference = 60

Simplifying the equation:
first term + 29 * common difference = 60

Now, we have a system of equations:
first term + 5 * common difference = 2 * (first term + 2 * common difference)
first term + 29 * common difference = 60

Let's solve this system of equations using substitution method:

From Equation 1, we can express the first term in terms of the common difference:
first term = 2 * first term + 4 * common difference - 5 * common difference
first term = - common difference

Substituting this value into Equation 2, we have:
- common difference + 29 * common difference = 60
28 * common difference = 60
common difference = 60 / 28
common difference ≈ 2.143

So, the common difference of the arithmetic sequence is approximately 2.143.

Sure, I can help you with these problems. Let's go through each one step by step.

1. To find the perimeter and area of the seventh square, we need to first understand the pattern of forming new squares.

Given that we start with a square with a side length of 2 decimeters, we join the midpoints of each side to form a new square. This process is repeated 6 times.

To find the side length of the seventh square, we can observe that each time we perform this procedure, the side length of the new square becomes half of the previous square. So, the side length of the seventh square would be 2 decimeters divided by 2 raised to the power of 6 (since the procedure is repeated 6 times).

2. To find the numbers between 15 and 810 that have 8 as their units digit, we can follow these steps:

Start by determining the first number between 15 and 810 that has 8 as its units digit. In this case, it would be 18.

Next, determine the last number between 15 and 810 that has 8 as its units digit. In this case, it would be 808.

Now, we need to find the number of terms in this sequence. In an arithmetic sequence, we can use the formula: number of terms = (last term - first term) / common difference + 1. Since the common difference is 10 (due to the units digit being 8), the number of terms would be (808 - 18) / 10 + 1.

Finally, we can find the sum of the numbers in this sequence using the formula for the sum of an arithmetic series: sum = (number of terms / 2) * (first term + last term).

3. To find the common difference in an arithmetic sequence, given that the sixth term is twice the 3rd term and the 30th term is 60, we can use the following steps:

Let's assume the first term is "a" and the common difference is "d".

Since the sixth term is twice the 3rd term, we can write the equation: a + 5d = 2(a + 2d).

Similarly, since the 30th term is given as 60, we can write the equation: a + 29d = 60.

Now, we have a system of two equations with two variables. We can solve this system of equations simultaneously to find the values of "a" and "d". Once we have the value of "d", we will have found the common difference.

I hope this helps! Let me know if you have any further questions.