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3 years ago

A farmer planted corn in a square field. one side of the field measures 32 yards. what is the area of the corn field

Mathematics

1 answer:

UkoKoshka [18]3 years ago

7 0

The area is 1024 sq yards

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The hanger image below represents a balanced equation.

cricket20 [7]

Answer:

$\frac{1}{5} + z = \frac{3}{5}$

5 0

2 years ago

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I don’t know where to start with this problem

elixir [45]

Answer:

√(4/5)

Step-by-step explanation:

First, let's use reflection property to find tan θ.

tan(-θ) = 1/2

-tan θ = 1/2

tan θ = -1/2

Since tan θ < 0 and sec θ > 0, θ must be in the fourth quadrant.

Now let's look at the problem we need to solve:

sin(5π/2 + θ)

Use angle sum formula:

sin(5π/2) cos θ + sin θ cos(5π/2)

Sine and cosine have periods of 2π, so:

sin(π/2) cos θ + sin θ cos(π/2)

Evaluate:

(1) cos θ + sin θ (0)

cos θ

We need to write this in terms of tan θ. We can use Pythagorean identity:

1 + tan² θ = sec² θ

1 + tan² θ = (1 / cos θ)²

±√(1 + tan² θ) = 1 / cos θ

cos θ = ±1 / √(1 + tan² θ)

Plugging in:

cos θ = ±1 / √(1 + (-1/2)²)

cos θ = ±1 / √(1 + 1/4)

cos θ = ±1 / √(5/4)

cos θ = ±√(4/5)

Since θ is in the fourth quadrant, cos θ > 0. So:

cos θ = √(4/5)

Or, written in proper form:

cos θ = (2√5) / 5

4 0

3 years ago

A concentric-tube heat exchanger cools hot water owing at 1 kg/s from 80C to 30C. It uses air owing at 5 kg/s to perform this co

nordsb [41]

Answer:

The log-mean-temperature-difference is 24.03⁰C

Step-by-step explanation:

First we need to know if the heat exchanger is in parallel flow or counter-flow. However, counter flow arrangement is best used to recover heat.

L.M.T.D for counter flow is given as;

$L.M.T.D =\frac{(T_h_f_1 -T_c_f_2)-(T_h_f_2 -T_c_f_1)}{2.3log[\frac{T_h_f_1 -T_c_f_2}{T_h_f_2 -T_c_f_1}]}$

where;

Thf₁ is the initial temperature of the hot fluid = 80°C

Tcf₂ is the final temperature of the cold fluid = 51.5°C

Thf₁ - Tcf₂ = 80 - 51.5 = 28.5⁰C

Thf₂ is the final temperature of the hot fluid = 30°C

Tcf₁ is the initial temperature of the cold fluid = 10°C

Thf₂ - Tcf₁ = 30 - 10 = 20⁰C

$L.M.T.D = \frac{28.5 -20}{2.3Log[\frac{28.5}{20}]} \\\\L.M.T.D = \frac{8.5}{0.3538} =24.03^oC$

Therefore, the log-mean-temperature-difference is 24.03⁰C

3 0

2 years ago

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2 friends share 3 fruit bars equally. How much does each friend get?

gizmo_the_mogwai [7]

1.5 for each friend, plus who eats fruit bars xD

5 0

3 years ago

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PLEASE HELP ASAP!!!!!

dezoksy [38]

I think 4 most likely because if u take the middle of the 3 dots it would lave the last dot on the 4

4 0

2 years ago