Solve for <span>xx</span> by simplifying both sides of the equation, then isolating the variable.<span>x=<span>34.62</span></span>
1. The shape of cross-section is a circle.
2. The face parallel to ABCD is EFGH. Since this is a a rectangular shape,
A = L*H = 12*6 = 72 cm^2
3. The cross-section parallel to ABC is DEF with h = 12 ft, b= 5ft (where h is the height and b is the base of a right angled triangle).
Area, A = 1/2 *b*h = 1/2*5*12 =30 ft^2
4. Plane BDHF is a rectangle shape whose length is the diagonal of ABCD.
Diagonal BD = sqrt (AB^2+BD^2) = sqrt (8^2+7^2) = 10.63 cm.
Perimeter, P = 2(BD+DH) = 2(10.63+6) = 33.26 cm
Answer:
C. 2
Step-by-step explanation:
h/6 + h/3 = 1
h/6 + 2h/6 = 1
3h/6 = 1
3h = 6
h = 2
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 area is 44.18 , hope this helps
