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vladimir1956 [14]

2 years ago

8

A farmer has 12 hectares of land on which he grows corn, wheat, and soybeans. It costs $4500 per hectare to grow corn, $6000 to

grow wheat, and $5000 to grow soybeans. Because of market demand the farmer will grow twice as many hectares of wheat as corn. He has allocated $6,375,000.00 for the cost of growing his crops. How many hectares of each crop should he plant?

1 answer:

maw [93]2 years ago

5 0

The number of hectares of each crop he should plant are; 250 hectares of Corn, 500 hectares of Wheat and 450 hectares of soybeans

<h3>How to solve algebra word problem?</h3>

He grows corn, wheat and soya beans on the farm of 1200 hectares. Thus;

C + W + S = 12 ----(1)

It costs $45 per hectare to grow corn, $60 to grow wheat, and $50 to grow soybeans. Thus;

45C + 60W + 50S = 63750 -----(2)

He will grow twice as many hectares of wheat as corn. Thus;

W = 2C ------(3)

Put 2C for W in eq 1 and eq 2 to get;

C + 2C + S = 1200

3C + S = 1200 -----(4)

45C + 60(2C) + 50S = 63750

45C + 120C + 50S = 63750

165C + 50S = 63750 ------(5)

Solving eq 4 and 5 simultaneosly gives;

C = 250 and W = 500

Thus; S = 1200 - 3(250)

S = 450

Read more about algebra word problems at; brainly.com/question/13818690

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Air enters a diffuser operating at steady state at 540°R, 15 lbf/in.2, with a velocity of 600 ft/s, and exits with a velocity of

yKpoI14uk [10]

Answer: Hello the question is incomplete below is the missing part

Question: determine the temperature, in °R, at the exit

answer:

T2= 569.62°R

Explanation:

T1 = 540°R

V2 = 600 ft/s

V1 = 60 ft/s

h1 = 129.0613 ( value gotten from Ideal gas property-air table )

<em>first step : calculate the value of h2 using the equation below </em>

assuming no work is done ( potential energy is ignored )

h2 = [ h1 + ( V2^2 - V1^2 ) / 2 ] * 1 / 32.2 * 1 / 778

∴ h2 = 136.17 Btu/Ibm

From Table A-17

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attached below is the remaining part of the solution

8 0

3 years ago

Find the power and the rms value of the following signal square: x(t) = 10 sin(10t) sin(15t)

ArbitrLikvidat [17]

Answer:

$\mathbf{P_x =25 \ watts}$

$\mathbf{x_{rmx} = 5 \ unit}$

Explanation:

Given that:

x(t) = 10 sin(10t) . sin (15t)

the objective is to find the power and the rms value of the following signal square.

Recall that:

sin (A + B) + sin(A - B) = 2 sin A.cos B

x(t) = 10 sin(15t) . cos (10t)

x(t) = 5(2 sin (15t). cos (10t))

x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)

x(t) = 5sin(25 t) + 5 sin (5t)

From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:

$P= \dfrac{A^2}{2}$

For the number of sinosoidial signals;

Power can be expressed as:

$P = \dfrac{A_1^2}{2}+ \dfrac{A_2^2}{2}+ \dfrac{A_3^2}{2}+ ...$

As such,

For x(t), Power $P_x = \dfrac{5^2}{2}+ \dfrac{5^2}{2}$

$P_x = \dfrac{25}{2}+ \dfrac{25}{2}$

$P_x = \dfrac{50}{2}$

$\mathbf{P_x =25 \ watts}$

For the number of sinosoidial signals;

$RMS = \sqrt{(\dfrac{A_1}{\sqrt{2}})^2+(\dfrac{A_2}{\sqrt{2}})^2+(\dfrac{A_3}{\sqrt{2}})^2+...$

For x(t), the RMS value is as follows:

$x_{rmx} =\sqrt{(\dfrac{5}{\sqrt{2}} )^2 +(\dfrac{5}{\sqrt{2}} )^2 }$

$x_{rmx }=\sqrt{(\dfrac{25}{2} ) +(\dfrac{25}{2} ) }$

$x_{rmx }=\sqrt{(\dfrac{50}{2} )}$

$x_{rmx} =\sqrt{25}$

$\mathbf{x_{rmx} = 5 \ unit}$

8 0

3 years ago

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True .the chlorofluorocarbons also known as CFC’s also contribute on depleting the ozone layer

8 0

3 years ago

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andriy [413]

The law requires employers to provide their employees with working conditions that are free of known dangers. The OSH Act created the Occupational Safety and Health Administration (OSHA), which sets and enforces protective workplace safety and health standards.

6 0

4 years ago

A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal

Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

$\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)$

$10*2=\sqrt{1+8F^{2} } -1$

Solving for F:

F = 7.4162

$v=F\sqrt{gd_{1} }$

Here, g = gravity = 32.2 ft/s²

$v=7.4162*\sqrt{32.2*1} =42.0833ft/s$

b) The flow rate:

$q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s$

c) The Froude number:

$v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s$

$F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345$

d) The flow energy dissipated:

$E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft$

e) The critical depth:

$d_{c} =(\frac{(\frac{q}{L})^{2} }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2} }{32.2} )^{1/3} =3.8030ft$

4 0

4 years ago