Answer:
<h2>54m²</h2>
Step-by-step explanation:
<h3>METHOD 1:</h3>
You can use the Heron's formula:
![A=\sqrt{p(p-a)(p-b)(p-c)}](https://tex.z-dn.net/?f=A%3D%5Csqrt%7Bp%28p-a%29%28p-b%29%28p-c%29%7D)
where
<em>p</em><em> - half of perimeter</em>
<em>a, b, c</em><em> - lengths of sides</em>
We have
![a=12m;\ b=15m;\ c=9m](https://tex.z-dn.net/?f=a%3D12m%3B%5C%20b%3D15m%3B%5C%20c%3D9m)
Calculate:
![p=\dfrac{12+15+9}{2}=\dfrac{36}{2}=18\ (m)\\\\A=\sqrt{18(18-12)+(18-15)(18-9)}\\\\A=\sqrt{(18)(6)(3)(9)}\\\\A=\sqrt{2916}\\\\A=54\ (m^2)](https://tex.z-dn.net/?f=p%3D%5Cdfrac%7B12%2B15%2B9%7D%7B2%7D%3D%5Cdfrac%7B36%7D%7B2%7D%3D18%5C%20%28m%29%5C%5C%5C%5CA%3D%5Csqrt%7B18%2818-12%29%2B%2818-15%29%2818-9%29%7D%5C%5C%5C%5CA%3D%5Csqrt%7B%2818%29%286%29%283%29%289%29%7D%5C%5C%5C%5CA%3D%5Csqrt%7B2916%7D%5C%5C%5C%5CA%3D54%5C%20%28m%5E2%29)
<h3>METHOD 2:</h3>
Let's check that it is not a right triangle.
If the sum of the squares of the two shorter sides is equal to the square of the longest side, then this triangle is rectangular.
We have
![9m < 12m](https://tex.z-dn.net/?f=9m%20%3C%2012m%3C15m)
Check:
![9^2+12^2=81+144=225\\15^2=225](https://tex.z-dn.net/?f=9%5E2%2B12%5E2%3D81%2B144%3D225%5C%5C15%5E2%3D225)
This is a right trianglr wherew 9m and 12m are legs and 15m is a hypotenuse.
The formula of an area of a right triangle is:
![A=\dfrac{ab}{2}](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7Bab%7D%7B2%7D)
<em>a, b</em><em> - legs</em>
Substitute:
![A=\dfrac{(9)(12)}{2}=\dfrac{108}{2}=54\ (m^2)](https://tex.z-dn.net/?f=A%3D%5Cdfrac%7B%289%29%2812%29%7D%7B2%7D%3D%5Cdfrac%7B108%7D%7B2%7D%3D54%5C%20%28m%5E2%29)
Multiply it out.
I believe the answer would be 15x^2 - 20x.
However, functions is not my forte.
Soooo
yeah
Divde it by 12^2 or 144
1250/144=8.605555555555ft^2
Answer:
8 pieces of candy
Step-by-step explanation:
Take the original amount of candy in the bowl to be
.
It says that 20 pieces of candy were added, therefore there are
pieces of candy in the bowl.
Half of the candy is gone, therefore
pieces of candy are left in the bowl.
After half of the candy is gone, there are 14 pieces left, so we can say that ![\frac{x+20}{2} =14](https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B20%7D%7B2%7D%20%3D14)
Solve for
:
![x+20 = 28](https://tex.z-dn.net/?f=x%2B20%20%3D%2028)
![x=8](https://tex.z-dn.net/?f=x%3D8)