Check the picture below.
so the shape is really 4 triangles with a base of 2 and a height of 4 each, and 2 squares tha are 4x4.
![\bf \stackrel{\textit{area of the 4 triangles}}{4\left[\cfrac{1}{2}(2)(4) \right]}~~+~~\stackrel{\textit{area of the two squares}}{2(4\cdot 4)}\implies 16+32\implies 48](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%204%20triangles%7D%7D%7B4%5Cleft%5B%5Ccfrac%7B1%7D%7B2%7D%282%29%284%29%20%5Cright%5D%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20two%20squares%7D%7D%7B2%284%5Ccdot%204%29%7D%5Cimplies%2016%2B32%5Cimplies%2048)
Answer:
j=6
Step-by-step explanation:
2=5j-28
+28 +28
30=5j
/5 /5
j=6
Answer:
square
triangle
5
square pyramid
find area of base and area of a lateral face (base is 1 side squared, lateral face is base * height / 2. Multiply area of lateral face by 4 and add the product to the area of the base to find the Surface Area.
Step-by-step explanation:
Answer:
Step-by-step explanation:
16p² = 29
p² = 29/16
p = ±√(29/16)
x² = -18
x = ±√(-18)
x = ± √9•√2•√-1
x = ± 3√2i
5(x - 1)² + 42 = 447
5(x² - 2x + 1) + 42 = 447
5x² - 10x + 5 + 42 = 447
5x² - 10x - 400 = 0
x² - 2x - 80 = 0
x = (2 ±√(2² - 4(1)(-80))) / (2(1))
x = (2 ±√324) / 2
x = (2 ±18) / 2
x = 1 ± 9
x = -8 , 10
3(x - 9)² = -100
3(x² - 18x + 81) = -100
3x² - 54x + 243 = -100
3x² - 54x + 343 = 0
x = (54 ±√(54² - 4(3)(343))) / (2(3))
x = (54 ±√-1200) / 6
x = (54 ± (20√3)i) / 6
x = 9 ± ⅓(10√3)i
