Answer:
The radius of the inflated spherical balloon is 45 millimeters.
Step-by-step explanation:
Volume of the spherical water balloon = 121,500 pi cubic millimeters
Let the radius of the balloon = r
Now, Volume of a Sphere = 
⇒
On solving for the value of r, we get:
![r^{3} = \frac{121,500\times 3}{4} = 91125\\ \implies r = \sqrt[3]{91125}](https://tex.z-dn.net/?f=r%5E%7B3%7D%20%20%3D%20%5Cfrac%7B121%2C500%5Ctimes%203%7D%7B4%7D%20%20%20%3D%2091125%5C%5C%20%5Cimplies%20%20r%20%20%3D%20%5Csqrt%5B3%5D%7B91125%7D)
or, r = 45 millimeter
Hence, the radius of the inflated spherical balloon is 45 millimeters.
The length of the rope in this case becomes the radius of the circle that the goat is able to graze. The area of the circle is calculated through the equation,
A = πr²
Substituting,
A = π(5m)² = 78.54 m²
Thus, the area that the goat is able to graze is equal to 78.54 m².
Answer:
Solve for y in the second equation
Step-by-step explanation:
We assume your system is ...
Dividing <em>the </em><em>second equation</em> by 3 gives ...
-3x +y = 1
so an expression for y without fractions can be found by <em>solving for y</em>, that is, by adding 3x:
y = 3x +1
_____
<em>Comment on alternate solution</em>
We'd be tempted to solve the first equation for -9x and substitute for that.
-9x = 10 -4y
(10 -4y) +3y = 3 . . . . . substitute for -9x
-y = -7 . . . . . . . . . . . . . simplify, subtract 10
y = 7 . . . . . . . . . . . . . . multiply by -1
