<span>Simplifying
2x + 18y = 36
Solving
2x + 18y = 36
Solving for variable 'x'.
Move all terms containing x to the left, all other terms to the right.
Add '-18y' to each side of the equation.
2x + 18y + -18y = 36 + -18y
Combine like terms: 18y + -18y = 0
2x + 0 = 36 + -18y
2x = 36 + -18y
Divide each side by '2'.
x = 18 + -9y
Simplifying
x = 18 + -9y</span>
Answer:
Step-by-step explanation:
x is greater or equal to 13
Answer:
1) 250 meters= 0.25 km
2) 12 meters= 0.012km
3) 1 meter= 0.001 km
The constant of proportionality is 1000 meters in 1km (1000 meters)
I think there’s a mistake on the second part because 1 meter does not equal to 1000km. but I did part 1 for you!
Answer:
![\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}](https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%20%5Cpi%20%5Cquad%20or%20%5Cquad%20%5Cdfrac%7B7%7D%7B9%7D)
Step-by-step explanation:
The <u>width</u> of a square is its <u>side length</u>.
The <u>width</u> of a circle is its <u>diameter</u>.
Therefore, the largest possible circle that can be cut out from a square is a circle whose <u>diameter</u> is <u>equal in length</u> to the <u>side length</u> of the square.
<u>Formulas</u>
![\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}](https://tex.z-dn.net/?f=%5Csf%20%5Ctextsf%7BArea%20of%20a%20square%7D%3Ds%5E2%20%5Cquad%20%5Ctextsf%7B%28where%20s%20is%20the%20side%20length%29%7D)
![\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}](https://tex.z-dn.net/?f=%5Csf%20%5Ctextsf%7BArea%20of%20a%20circle%7D%3D%5Cpi%20r%5E2%20%5Cquad%20%5Ctextsf%7B%28where%20r%20is%20the%20radius%29%7D)
![\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}](https://tex.z-dn.net/?f=%5Csf%20%5Ctextsf%7BRadius%20of%20a%20circle%7D%3D%5Cdfrac%7B1%7D%7B2%7Dd%20%5Cquad%20%5Ctextsf%7B%28where%20d%20is%20the%20diameter%29%7D)
If the diameter is equal to the side length of the square, then:
![\implies \sf r=\dfrac{1}{2}s](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20r%3D%5Cdfrac%7B1%7D%7B2%7Ds)
Therefore:
![\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cimplies%20%5Csf%20Area%5C%3Aof%5C%3Acircle%20%26%20%3D%20%5Csf%20%5Cpi%20%5Cleft%28%5Cdfrac%7Bs%7D%7B2%7D%5Cright%29%5E2%5C%5C%26%20%3D%20%5Csf%20%5Cpi%20%5Cleft%28%5Cdfrac%7Bs%5E2%7D%7B4%7D%5Cright%29%5C%5C%26%20%3D%20%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%5Cpi%20s%5E2%20%5Cend%7Baligned%7D)
So the ratio of the area of the circle to the original square is:
![\begin{aligned}\textsf{area of circle} & :\textsf{area of square}\\\sf \dfrac{1}{4}\pi s^2 & : \sf s^2\\\sf \dfrac{1}{4}\pi & : 1\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Ctextsf%7Barea%20of%20circle%7D%20%26%20%3A%5Ctextsf%7Barea%20of%20square%7D%5C%5C%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%5Cpi%20s%5E2%20%26%20%3A%20%5Csf%20s%5E2%5C%5C%5Csf%20%5Cdfrac%7B1%7D%7B4%7D%5Cpi%20%26%20%3A%201%5Cend%7Baligned%7D)
Given:
- side length (s) = 6 in
- radius (r) = 6 ÷ 2 = 3 in
![\implies \sf \textsf{Area of square}=6^2=36\:in^2](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Ctextsf%7BArea%20of%20square%7D%3D6%5E2%3D36%5C%3Ain%5E2)
![\implies \sf \textsf{Area of circle}=\pi \cdot 3^2=28\:in^2\:\:(nearest\:whole\:number)](https://tex.z-dn.net/?f=%5Cimplies%20%5Csf%20%5Ctextsf%7BArea%20of%20circle%7D%3D%5Cpi%20%5Ccdot%203%5E2%3D28%5C%3Ain%5E2%5C%3A%5C%3A%28nearest%5C%3Awhole%5C%3Anumber%29)
Ratio of circle to square:
![\implies \dfrac{28}{36}=\dfrac{7}{9}](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B28%7D%7B36%7D%3D%5Cdfrac%7B7%7D%7B9%7D)
We know that the ocean floor has a depth of 247 ft, and we also know that the diver is<span> underwater at depth of 138 ft, so its distance from the ocean floor will be:
</span>
![247-138=109](https://tex.z-dn.net/?f=247-138%3D109)
ft
<span>
Now, the </span>rock formations rises to a peak 171 to above the ocean floor, so to find <span>how many feet below the top of the rock formations is the diver, we are going to subtract the distance to the driver form the ocean floor from the rock formations height:
</span>
![171-109=62](https://tex.z-dn.net/?f=171-109%3D62)
ft
<span>
We can conclude that the diver is 62 feet </span><span>
below the top of the rock formations.</span>