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GaryK [48]
3 years ago
11

Photon Lighting Company determines that the supply and demand functions for its most popular lamp are as follows: S(p) = 400 - 4

p + 0.00002p4 and D(p) = 2800 - 0.0012p3, where p is the price. Determine the price for which the supply equals the demand.
A) $93.24
B) $100.24
C) $96.24
D) $99.24

Physics
2 answers:
torisob [31]3 years ago
6 0

Answer: The correct answer is (C).

Explanation:

Supply function:

S(p)=400-4p+0.00002p^4

Demand function:

D(p)=2800-0.0012p^3

S(p)=D(p), price for which supply is equal to demand:

S(p)=D(p)=400-4p+0.00002p^4=2800-0.0012p^3

After solving this above equation graphically, we will get the values of p.

1) p =96.236

2) p=( -118.258)

We will reject the negative value of p.

So, the value of p that price for which the supply equals the demand is $ 96.236 \approx $96.24. Hence, the correct answer is option (C).

andrey2020 [161]3 years ago
3 0

Answer:

Price equals $96.24

Explanation:

It is given that,

The supply function is given by, S(p)=400-4p+0.00002p^4

The demand function is given by, D(p)=2800-0.0012p^3

We have to find the price for which the supply equals to the demand.

S(p)=D(p)

400-4p+0.00002p^4=2800-0.0012p^3

After solving above equation we get :

So, it is clear from the attached graph that p = 96.24 and q = -118.26

So, the correct option is (c) " $96.24 ".

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irina1246 [14]
Perimeter = 2 ( L + W )
32 = 2 ( L + W )
16 = L + W
L = 16 - W

Area = L W
63 = L W
63 = (16-W) W
63 = 16W - W²
-W² + 16 W - 63 = 0
By factorizing W = 9 or W = 7
So the dimensions are 7 and 9
3 0
3 years ago
A snowmobile has an initial velocity of 4.7 m/s.
zimovet [89]
Q1. The answer is 8.788 m/s

V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time

We have:
V1 = 4.7 m/s
a = 0.73 m/s²
t = 5.6 s
V2 = ?

V2 = 4.7 + 0.73 * 5.6
V2 = 4.7 + 4.088
V2 = 8.788 m/s


Q2. The answer is 9.22 s

V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time

We have:
V2 = 0   (because it reaches a complete stop)
V1 = 4.7 m/s
a = -0.51 m/s²
t = ?

0 = 4.7 + (-0.51)*t
0 = 4.7 - 0.51t
0.51t = 4.7
t = 4.7 / 0.51
t = 9.22 s
3 0
3 years ago
Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

3 0
3 years ago
12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
14. Which one of the following pictures shows the object that is the most dense? *
rewona [7]

Answer:

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