Answer:
Step-by-step explanation:
∵ The volume of the pyramid = 1/3 base area × height
∵ The base is equilateral Δ with side length 4
∴ The area of the bast = 1/4 × 4² × √3 = 4√3 units²
To get the height of the pyramid draw it from the vertex of the top of the pyramid ⊥ to the base on the centro-id of the base which divides the height of the triangle two ratio 2:1 from the vertex of the triangle
∵ The height of the base = √(4² - 2²) =√12 = 2√3
∴ 2/3 the height = 4√3/3 ⇒ (2:1 means 2/3 from the height)
∴ The height of the pyramid = √[4² - (4√3/3)²] = √[16 - 48/9]
∴ h = 4√2/√3 (4√6/3 in its simplest form)
∴ V = 1/3 × 4√3 × 4√2/√3 = 16√2/3 units³
∴ 
Answer:
The line between the points C and E is parallel to the line between the points F and B
The line between the points A and E is parallel to the line between the points B and D
The line between the points A and C is parallel to the line between the points F and D
Answer:
1. y' = 3x² / 4y²
2. y'' = 3x/8y⁵[(4y³ – 3x³)]
Step-by-step explanation:
From the question given above, the following data were obtained:
3x³ – 4y³ = 4
y' =?
y'' =?
1. Determination of y'
To obtain y', we simply defferentiate the expression ones. This can be obtained as follow:
3x³ – 4y³ = 4
Differentiate
9x² – 12y²dy/dx = 0
Rearrange
12y²dy/dx = 9x²
Divide both side by 12y²
dy/dx = 9x² / 12y²
dy/dx = 3x² / 4y²
y' = 3x² / 4y²
2. Determination of y''
To obtain y'', we simply defferentiate above expression i.e y' = 3x² / 4y². This can be obtained as follow:
3x² / 4y²
Let:
u = 3x²
v = 4y²
Find u' and v'
u' = 6x
v' = 8ydy/dx
Applying quotient rule
y'' = [vu' – uv'] / v²
y'' = [4y²(6x) – 3x²(8ydy/dx)] / (4y²)²
y'' = [24xy² – 24x²ydy/dx] / 16y⁴
Recall:
dy/dx = 3x² / 4y²
y'' = [24xy² – 24x²y (3x² / 4y² )] / 16y⁴
y'' = [24xy² – 18x⁴/y] / 16y⁴
y'' = 1/16y⁴[24xy² – 18x⁴/y]
y'' = 1/16y⁴[(24xy³ – 18x⁴)/y]
y'' = 1/16y⁵[(24xy³ – 18x⁴)]
y'' = 6x/16y⁵[(4y³ – 3x³)]
y'' = 3x/8y⁵[(4y³ – 3x³)]
The answer is 669
5^4-2(5)^3+5(5)^2+-7(5)+4
625-250+125-35+4
hope this helps!
