The area of a rectangle is Base x height whereas the area of a triangle is Base x height divided by 2
Answer:
<h2>Answer :</h2>
The original price is x, so 20% of x, - x, equals the sale price. You can use the equation:
x-.2x=15.20
.8x=15.20
/.8. /.8
x=19
have a nice day <3
Step-by-step explanation:
Answer:
you can use math way it will be on there
Answer:
Step-by-step explanation:
Given
Two curves are given

and 
the two curves intersect at


to get the we need to integrate the curves over x axis


![A=2\left [ 3\left ( \frac{\sqrt{2}}{3}\right )-\frac{9}{3}\left ( \frac{\sqrt{2}}{3}\right )^3\right ]](https://tex.z-dn.net/?f=A%3D2%5Cleft%20%5B%203%5Cleft%20%28%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B3%7D%5Cright%20%29-%5Cfrac%7B9%7D%7B3%7D%5Cleft%20%28%20%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B3%7D%5Cright%20%29%5E3%5Cright%20%5D)
![A=2\sqrt{2}\left [ 1-\frac{2}{9}\right ]](https://tex.z-dn.net/?f=A%3D2%5Csqrt%7B2%7D%5Cleft%20%5B%201-%5Cfrac%7B2%7D%7B9%7D%5Cright%20%5D)

Answer:
-2,4/3
Step-by-step explanation:
ax²+bx+c=0
