Two cuboids have the same surface area
1 answer:
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Answer:
The length of DC in meters is ⇒ A
Step-by-step explanation:
In the circle O
∵ AB passing through O
∴ AB is a diameter
∵ D is on the circle
∴ ∠ADB is an inscribed angle subtended by arc AB
∵ Arc AB is half the circle
→ That means its measure is 180°
∴ m∠ADB = × 180° = 90°
In ΔADB
∵ m∠ADB = 90°
∵ AD = 5 m
∵ BD = 12 m
→ By using Pythagorase Theorem
∵ (AB)² = (AD)² + (DB)²
∴ (AB)² = (5)² + (12)²
∴ (AB)² = 25 + 144 = 169
→ Take square root for both sides
∴ AB = 13 m
∵ ∠ADB is a right angle
∵ DC ⊥ AB
∴ DC × AB = AD × DB
→ Substitute the lengths of AB, AD, and DB
∵ DC × 13 = 5 × 12
∴ 13 DC = 60
→ Divide both sides by 13
∴ DC = m
∴ The length of DC in meters is
Answer:
See below.
Step-by-step explanation:
Are f(x)=x2 and g(x) =4/5x2 correctly written?
This answer assumes they are supposed to be:
f(x)=x^2 and g(x) = (4/5)x^2
If these revisions are correct, g(x) would produce values that are (4/5) of f(x).
See attached graph.
I don't see any of the options for "best statement."
The given displacement at time, t, is
p(t) = 2 sin(t³) + 5 cos(t³)
The initial equilibrium position is
p(0) = 5
To determine future equilibrium postions, define
f(t) = p(t) - 5 = 2 sin(t³) + 5 cos(t³ - 5
The derivative of f(t) is
f'(t) = (3t²)[2 cos(t³) - 5sin(t³)]
Equilibrium is established when f(t) = 0.
To solve this equation numerically, we shall use the Newton-Raphson method, given by.
t(n+1) = t(n) - f[t(n)]/f'[(t(n)], n=0,1,2, ...,
As a guess, let (0) = 1.
The iterative solution for t is shown below.
n t(n)
--- -----------
0 1.0000
1 0.9344
2 0.9147
3 0.9130
4 0.9130
The solution converges rapidly to t = 0.913 s.
The graphical solution (shown below) confirms the numerical solution.
Answer:
The weight first reaches the equilibrium position in 0.913 sec.
p(t) = 2 sin(t³) + 5 cos(t³)
The initial equilibrium position is
p(0) = 5
To determine future equilibrium postions, define
f(t) = p(t) - 5 = 2 sin(t³) + 5 cos(t³ - 5
The derivative of f(t) is
f'(t) = (3t²)[2 cos(t³) - 5sin(t³)]
Equilibrium is established when f(t) = 0.
To solve this equation numerically, we shall use the Newton-Raphson method, given by.
t(n+1) = t(n) - f[t(n)]/f'[(t(n)], n=0,1,2, ...,
As a guess, let (0) = 1.
The iterative solution for t is shown below.
n t(n)
--- -----------
0 1.0000
1 0.9344
2 0.9147
3 0.9130
4 0.9130
The solution converges rapidly to t = 0.913 s.
The graphical solution (shown below) confirms the numerical solution.
Answer:
The weight first reaches the equilibrium position in 0.913 sec.