Answer:
The answer to the question is 5 (positive 5)
Answer:
![\large\boxed{C.\ (25+25\sqrt3)\ in^2}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7BC.%5C%20%2825%2B25%5Csqrt3%29%5C%20in%5E2%7D)
Step-by-step explanation:
We have the square and four equilateral triangles.
The formula of an area of a squre:
![A_S=a^2](https://tex.z-dn.net/?f=A_S%3Da%5E2)
a - length of side
The formula of an area of an equilateral triangle:
![A_T=\dfrac{a^2\sqrt3}{4}](https://tex.z-dn.net/?f=A_T%3D%5Cdfrac%7Ba%5E2%5Csqrt3%7D%7B4%7D)
a - length of side
Clculate the areas:
SQURE:
![A_S=x^2\ in^2](https://tex.z-dn.net/?f=A_S%3Dx%5E2%5C%20in%5E2)
TRIANGLE:
![A_T=\dfrac{x^2\sqrt3}{4}\ in^2](https://tex.z-dn.net/?f=A_T%3D%5Cdfrac%7Bx%5E2%5Csqrt3%7D%7B4%7D%5C%20in%5E2)
The SURFACE AREA of a square pyramid:
![S.A.=A_S+4A_T\\\\S.A.=x^2+4\cdot\dfrac{x^2\sqrt3}{4}=x^2+x^2\sqrt3](https://tex.z-dn.net/?f=S.A.%3DA_S%2B4A_T%5C%5C%5C%5CS.A.%3Dx%5E2%2B4%5Ccdot%5Cdfrac%7Bx%5E2%5Csqrt3%7D%7B4%7D%3Dx%5E2%2Bx%5E2%5Csqrt3)
Put x = 5:
![S.A.=5^2+5^2\sqrt3=(25+25\sqrt3)\ in^2](https://tex.z-dn.net/?f=S.A.%3D5%5E2%2B5%5E2%5Csqrt3%3D%2825%2B25%5Csqrt3%29%5C%20in%5E2)
To get the average, you must add all together and divide by the amount of numbers there are
(-4 - 6 - 3 - 1 + 0 + 2 - 1 - 3)/ 8
(-16)/8 = -2
-2 is your average
hope this helps
10 and 8 over 25 I just guess
Answer:-5-6
Step-by-step explanation: