6pints>60 fluid ounces
5quarts= 20cups
Please find attached photograph for your answer. Please do mark me Brainliest
Answer:2 in.
Step-by-step explanation:
Given
Dimension of photo frame is ![20\ in.\times 24\ in.](https://tex.z-dn.net/?f=20%5C%20in.%5Ctimes%2024%5C%20in.)
If the photo cover an area of ![320\ in.^2](https://tex.z-dn.net/?f=320%5C%20in.%5E2)
Suppose x be the width of border
Therefore dimension of frame without border is
![A'=(20-2x)(24-2x)](https://tex.z-dn.net/?f=A%27%3D%2820-2x%29%2824-2x%29)
And
must be equal to ![320\ in.^2](https://tex.z-dn.net/?f=320%5C%20in.%5E2)
So,
![\Rightarrow (20-2x)(24-2x)=320](https://tex.z-dn.net/?f=%5CRightarrow%20%2820-2x%29%2824-2x%29%3D320)
![\Rightarrow (10-x)(12-x)=80](https://tex.z-dn.net/?f=%5CRightarrow%20%2810-x%29%2812-x%29%3D80)
![\Rightarrow 120-10x-12x+x^2=80](https://tex.z-dn.net/?f=%5CRightarrow%20120-10x-12x%2Bx%5E2%3D80)
![\Rightarrow x^2-22x+40=0](https://tex.z-dn.net/?f=%5CRightarrow%20x%5E2-22x%2B40%3D0)
![\Rightarrow x^2-20x-2x+40=0](https://tex.z-dn.net/?f=%5CRightarrow%20x%5E2-20x-2x%2B40%3D0)
![\Rightarrow (x-2)(x-20)=0](https://tex.z-dn.net/?f=%5CRightarrow%20%28x-2%29%28x-20%29%3D0)
Thus there are two values of x out of which
is not valid because it is not feasible
thus width of border is ![x=2\ in.](https://tex.z-dn.net/?f=x%3D2%5C%20in.)
The answer is 51. This is because if you 15 is added then you multiply by the 10