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Anna [14]
3 years ago
5

Can someone help? please its an timed assignment.

Mathematics
1 answer:
tekilochka [14]3 years ago
3 0

Answer:

Well I only know that the rectangle is 224 because if you multiply 56 by 4 because there are 4 sides you get 224

Step-by-step explanation:

Brainliest sowy I only knew one

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Which is greater 0.205 or 205/1000
Burka [1]
Neither... 

205/1000 equals .205

So they are both the same.

Hope this helps -J
8 0
3 years ago
Perimeter of a square with side 4 square root of 5
Darya [45]

Answer:

16\sqrt{5}

Step-by-step explanation:

4\sqrt{5}+4\sqrt{5}+4\sqrt{5}+4\sqrt{5}

16\sqrt{5}

7 0
3 years ago
What is the change in temperature between -8°c and 3°c ?
Leokris [45]
11 degrees Celsius. Because -8 to 0 is 8. Then 0 to 3 is 3. So if you add the 3 and the 8 together, you get your answer 11 degrees. So 11 degrees Celsius is your change.
3 0
3 years ago
Im 21 assignments behind try your best <3
mestny [16]

Answer:

<em>.5, 3, 8.5</em>

Step-by-step explanation:

  1. Divide the minutes by 2 since 6 is half of 12.
  2. Half of 1 is .5 or \frac{1}{2}
  3. Half of 6 is 3
  4. Half of 17 is 8.5

If I am incorrect in my reasoning, please let me know so that I can plan better for my future answers. Have an amazing day.

3 0
3 years ago
Read 2 more answers
* Some amount of billiard balls were arranged in an equilateral triangle. And 5 balls were extra. When the same set of billiard
Agata [3.3K]

Lets say billiard balls are arranged in rows to form an equilateral triangle, then the first row consists of 1 ball, second row consists of 2 balls, and third row consists of 3 balls,  and so on. So there must be n balls in the n^{th} row.  

So, the total number of balls that forms the equilateral triangle with n rows is:  

1+2+3+4+5+....+n=\frac{n(n+1)}{2}

Let x_1 and x_2 be the total number of balls in the first and second arrangements respectively.  

Then,

x_1=\frac{n(n+1)}{2} +5

It has been said that there were 11 lesser balls in the second arrangement:  

x_2=\frac{1+(n+1)}{2} \times (n+1)-11=(n+1) \times \frac{(n+2)}{2} -11

Since, x_1=x_2

\frac{(n+1)}{2} \times n+5=\frac{(n+2)}{2} \times(n+1)-11

multiplying both the sides by 2

(n+1)\times n+10=(n+2)(n+1)-22

n+n^2=n^2+n+2n+2-22-10

2n=22+10-2

2n=30

n=15

Therefore,

x_1=\frac{(n+1)}{2}\times n+5=\frac{15+1}{2}  \times 15+5=125

So, there were 125 balls at the set.

4 0
3 years ago
Read 2 more answers
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