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Solve the following problems:

60) Atmospheric pressure \( P \) at altitude \( a \) is given by the formula:
\[ P = P_0 e^{-0.00005a} \]
where \( P_0 \) is the pressure at sea level. Assume \( P_0 = 14.7 \, \text{lb/in}^2 \) (pounds per square inch).

- Find the pressure at an altitude of 13,000 ft.
- At what altitude is the pressure \( 14.7 \, \text{lb/in}^2 \)?

Options:
A) \( 7.674 \, \text{lb/in}^2 \), 10 ft
B) \( 7.831 \, \text{lb/in}^2 \), 0 ft
C) \( 7.674 \, \text{lb/in}^2 \), 100 ft
D) \( 7.674 \, \text{lb/in}^2 \), 0 ft

61) Find the elasticity given the demand function:
\[ q = D(x) = \frac{x+8}{2x} \]

Options:
A) \( E(x) = -600x(x + 8)^2 \)
B) \( E(x) = +8 \)
C) \( E(x) = \)
D) \( E(x) = +8 \)

62) For the given demand function, find the value(s) of \( x \) for which total revenue is a maximum:
\[ x D(x) = 300 - 6x \]

Options:
A) 100
B) 50
C) 40
D) 25

63) Evaluate the integral:
\[ \int (8t^2 - 5t - 2) \, dt \]

Options:
A)
B) \( 8t^3 - 5t^2 - 2t + C \)
C) \( 16t - 5 + C \)
D) \( 4t^3 - 5t^2 - 2t + C \)

64) Evaluate:
\[ \int 31 \, dx \]

Options:
A)
B) \( 31x + C \)
C)
D) \(-31x + C\)

65) Find \( f \) such that the given conditions are satisfied:
Given \( f(x) = x^2 + 4 \), \( f(0) = 23 \)

Options:
A) \( f(x) = x^3 + 4x + 23 \)
B) \( f(x) = x^3 + 4x^2 + 23 \)
C) \( f(x) = +A + 23 \)
D) \( f(x) = x + 4x \)

Answer :

The value of pressure is 14.7 lb/in² at sea level. so correct option is A) The elasticity function E(x) is (x + 8) / (2x^2). so correct option is B). Total revenue R(x) is 25. so correct option is D). Integrating the given function 8t^2 - 5t - 2 gives (8/3)t^3 - (5/2)t^2 - 2t + C. so correct option is B). Integrating the constant function 31 gives 31x + C. so correct option is B). The value of F(x) is x² + 23. The correct option is C).

We are given the atmospheric pressure at an altitude a as P = Po - 0.00005a, where Po = 14.7 lb/in².

To find the pressure at an altitude of 13,000 ft, we convert the altitude to inches and substitute it into the equation for P:

Altitude = 13,000 ft = 13,000 * 12 in/ft = 156,000 in

P = Po - 0.00005a = 14.7 - 0.00005(156,000) = 6.9 lb/in²

Therefore, the pressure at an altitude of 13,000 ft is 6.9 lb/in².

To find the altitude at which the pressure is 14.7 lb/in², we set P = Po in the equation for P and solve for a:

P = Po - 0.00005a

14.7 = 14.7 - 0.00005a

0.00005a = 0

a = 0

Therefore, the pressure is always 14.7 lb/in² at sea level (altitude = 0 ft).

So, the correct answer is A).

The elasticity function E(x) is given by:

E(x) = D(x) / (x * P(x))

where D(x) is the demand function and P(x) is the price function.

Here, D(x) = x + 8 and P(x) = 2x.

Substituting these values, we get:

E(x) = (x + 8) / (2x^2)

Answer is B) E(x) = (x + 8) / (2x^2).

Total revenue R(x) is given by:

R(x) = x * D(x)

where D(x) is the demand function.

Here, D(x) = 300 - 6x.

Substituting these values, we get:

R(x) = x(300 - 6x) = 300x - 6x^2

Differentiating R(x) w.r.t x, we get:

dR/dx = 300 - 12x

Setting dR/dx = 0 to find the maximum, we get:

300 - 12x = 0

x = 25

Therefore, x = 25 is the value for which total revenue is a maximum.

Answer is D) 25.

Integrating the given function 8t^2 - 5t - 2 w.r.t t, we get:

∫(8t^2 - 5t - 2) dt = (8/3)t^3 - (5/2)t^2 - 2t + C

where C is the constant of integration.

Answer is B) (8/3)t^3 - (5/2)t^2 - 2t + C.

Integrating the constant function 31 w.r.t x, we get:

∫31 dx = 31x + C

where C is the constant of integration.

Answer is B) 31x + C.

We are given the function f(x) = x² + 4 and f(0) = 23.

To find the form of the function, we can integrate f'(x) = 2x to obtain f(x) = x² + C, where C is the constant of integration. We can then use the initial condition f(0) = 23 to find the value of C:

f(0) = 0² + C = C = 23

Therefore, f(x) = x² + 23.

The answer is (C) f(x) = x² + 23.

To know more about Integral:

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--The given question is incomplete, the complete question is given

" Solve. 60) Atmospheric pressure P at altitude a is given by P- Poe-0.00005a, where Po is the pressure at sea level. Assume that Po 14.7 lbjin2 (pounds per square inch). Find the pressure at an altitude of 13,000 ft. At what altitude is the pressure A) 14.7 lb/in?? B) 7.831 Ib/in2,0 ft D) 7.674 lbin2;0 ft A) 7.674 lb/in2, 10 ft C) 7.674 lb/in2; 100 ft Find the elasticity. 300

61) q= D(x)=(x+8,2 2x B) E(x)+8 A) E(x)- 600x(x + 8)2 B) E(x) = (x + 8) / (2x^2). D) E(x)+8 600x (x+ 8)3 C) E(x) = 8x^2 For the given demand function, find the valuets) of x for which total revenue is a B) 50 which total revenue is a maximum

62) x D(x)-300-6x A) 100 D) 25 C) 40

63) Evaluate. (8t2-5t-2) dt 63 B)8t3-5t2-2t + C 2-2t+ C D) 4t3-5t2-2t+C C)16t-5+C 31 dx

64) 31 31 A)- C D) -31x+ C B) 31x + C Find fsuch that the given conditions are satisfed.

65) f(x) = x2 + 4, f(0) = 23 A) f(x) x3+4x+23 C)f(x) = +A+23 B) fx) x3+4x2+23 D) ffx)-X+ 4x"--

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