So, (6a - X)*5a = Y*(a^2) - 35a
<span>=> 6a*5a - X*5a = Y*(a^2) - 35a </span>
<span>=> 30(a^2) - X*5a = Y(a^2) - 35a
</span>30 = Y and
<span>X*5 = 35 or X = 7 </span>
Answer:
about 51.77%
Step-by-step explanation:
The probability you will roll a 2 is the complement of the probability you will not roll a 2.
__
<h3>p(no 2)</h3>
A die has 6 faces, one of which is labeled with 2. The probability of rolling a 2 on any given roll is 1/6, so is 1-1/6 = 5/6 that a 2 will not be rolled. Sequential rolls of the die are presumed to be independent, so the probability of 4 rolls not showing a 2 at any time is ...
p(no 2) = (5/6)^4 = 625/1296
<h3>p(2)</h3>
The probability that at least one 2 will show up on at least one roll is the complement of this value:
p(some 2) = 1 -625/1296 = 671/1296 ≈ 0.517747
The chance you will roll at least one 2 is about 51.77%.
A rotation, because they are circles a rotation wouldn’t work.
Answer:
Perimeter: ![4x^2-2x+8](https://tex.z-dn.net/?f=4x%5E2-2x%2B8)
Area: ![8x^2-4x](https://tex.z-dn.net/?f=8x%5E2-4x)
Step-by-step explanation:
I will assume the question asks for perimeter and area of the rectangle with given dimensions.
The height is "4"
The width (assuming) is "
"
First, the perimeter:
Perimeter is sum of all 4 sides of a rectangle, so we have 2 widths and 2 lengths, we sum them up:
![Perimeter=4+2x^2-x+4+2x^2-x=4x^2-2x+8](https://tex.z-dn.net/?f=Perimeter%3D4%2B2x%5E2-x%2B4%2B2x%5E2-x%3D4x%5E2-2x%2B8)
Now, the area:
Area is the height times width, so it will be:
![Area=4*(2x^2-x)=8x^2-4x](https://tex.z-dn.net/?f=Area%3D4%2A%282x%5E2-x%29%3D8x%5E2-4x)
Not all lines are proportional and have a set ratio