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Arlecino [84]
2 years ago
6

How many squares would be in the 7th pattern?

Mathematics
2 answers:
Nitella [24]2 years ago
8 0

Answer: I believe in would be 28.

Explanation:

NeTakaya2 years ago
4 0

Answer:

I would say that there would be 28 squares in the 7th pattern.

Step-by-step explanation:

If you look at how the pattern is gradually increasing every time you add the number located at the bottom to the previous pattern. So if that is the rule then you would add <u>7 more squares</u> to the <u>21 squares</u> in the 6th pattern. Then you add

21 + 7 and you get 28.

<u>Ex</u>: In the first pattern there is 1 square and then in the 2nd pattern you see that there is an additional 2 squares which is the number at the bottom of the actual pattern.

*Idk if this is the actual rule to the pattern so please do not come for me if I am wrong but this is the way my brain sees the patterns in the picture. I also went ahead and tried my theory for all the patterns in the picture and it works :)

I really hope that this helps you understand the question a bit more! :)

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3 years ago
Solve for x. Show each step of the solution 4.5 ( 8 - x ) + 36 = 102 - 2.5 ( 3x + 24)​
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Answer:

X=-10

Step-by-step explanation:

3 0
3 years ago
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Together, two people earn $28,000. One earns $2,000 more than the other. How much is the smaller income?
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Look at this:

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For which triangle does the equation 42 + 42 = c2 apply?
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Step-by-step explanation:

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2 years ago
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Find the vector that has the same direction as 3, 2, −6 but has length 2.
Nata [24]

Answer:

The vector is \vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right).

Step-by-step explanation:

We can determine the equivalent vector (\vec r), dimensionless, by means of the following formula:

\vec r = \frac{\vec u}{\|\vec u\|} \cdot \|\vec r\| (1)

Where:

\vec u - Original vector, dimensionless.

\|\vec u\| - Norm of the original vector, dimensionless.

\|\vec r\| - Norm of the new vector, dimensionless.

The norm of the original vector is determined by the following definition:

\|\vec u\| = \sqrt{\vec u\,\bullet \,\vec u} (2)

If we know that \vec u = (3, 2, -6), then the norm of the original vector is:

\|\vec u\| = \sqrt{(3)\cdot (3)+(2)\cdot (2)+(-6)\cdot (-6)}

\|\vec u\| = 7

If we know that \|\vec r\| = 2, then the new vector is:

\vec r = \frac{2}{7}\cdot (3,2,-6)

\vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right)

The vector is \vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right).

7 0
3 years ago
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