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Natali5045456 [20]
3 years ago
12

A diet is to contain at least 2400 mg vitamin C, 1800mg Calcium, and 1200 calories every day. Two foods, a dairy-based meal and

a vegan option are to fulfill these requirements. Each ounce of the dairy-based meal provides 50 mg vitamin C, 30 mg Calcium, and 10 calories. Each ounce of the vegan option provides 20 mg vitamin C, 20 mg Calcium, and 40 calories. If the dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce, how many ounces of each food should be purchased to minimize costs? What is that minimum cost (per day)?
Physics
1 answer:
Kay [80]3 years ago
4 0

Answer:

The answer is below

Explanation:

Let x represent the number of ounce of dairy based meal and let y represent the number of vegan option in ounce.

Since the diet must contain at least 2400 mg vitamin C, therefore:

50x + 20y ≥ 2400

Since the diet must contain at least 1800 mg Calcium, therefore:

30x + 20y ≥ 1200

Since the diet must contain at least 1200 calories, therefore:

10x + 40y ≥ 1200

Therefore the constraints are:

50x + 20y ≥ 2400

30x + 20y ≥ 1200

10x + 40y ≥ 1200

x > 0, y > 0

The graph was drawn using geogebra online graphing tool, and the solution to the problem is at:

C(30, 45) and D(48, 18)

dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce. The cost equation is:

Cost = 0.042x + 0.208y

At C(30, 45);  Cost = 0.042(30) + 0.208(45) = $10.62

At C(48, 18);  Cost = 0.042(48) + 0.208(18) = $5.76

The minimum cost is at (48, 18). That is 48 dairy based meal and 18 vegan

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Answer:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m

Explanation:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m ok

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A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.36
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Answer: the average speed of the rat from the information given above is 0.7m/s

Explanation:

position is given as

x(t) = pt² + qt

finding the diffencial of x(t) with respect to t, we have

d(x(t))/dt = 2pt + q

we substitute the p = 0.36m/s² and q= -1.10 m/s

d(x(t))/dt = 2(0.36)t + (-1.10)

so, at t= 1s

d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s

at t= 4s

d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s

To find the average speed,

average speed = (V1 + V2)/ 2

average speed = (1.78 + (-0.38))/2 = 0.7m/s

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When the voltage is
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A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be
lora16 [44]

Answer:

frequency is 195.467 Hz

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given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × 10^{-3} m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36  

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ)       ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v  /  2     ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × 10^{-3})²×ω²× 34.64 )   /  2

0.050 ×(6.43 × 10^{-3})²×ω²× 34.64 = 108

ω² = 108 / 7.160  × 10^{-5}

ω = 1228.16 rad/s

so frequency will be

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7 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

3 0
2 years ago
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