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Naddika [18.5K]
3 years ago
12

What is 3/4 of a liter?

Mathematics
1 answer:
Rom4ik [11]3 years ago
3 0
3/4 of a liter is <span>0.75 liter I hope that helps!</span>
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Simplify: −6u2v−uv2−22u2v2
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Answer:

=-102uv

Step-by-step explanation:

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2. Evaluate the expression:<br> (5w^2+<br> +2) = (3p) for W= 4 and p = 3
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I have no clue on this one either
taurus [48]

Answer:

Area of the triangle = 210 cm^2 .

Step-by-step explanation:

Perimeter of triangle = 4x + 1 + 3x - 1 + 3x = 70

10x = 70 - 1 + 1 = 70

x = 7.

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= 1/2 * 3(7) * (3(7)  -1)

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= 210.

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2 years ago
HELPPPP PLSSS &lt;3 TYYy
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Step-by-step explanation:

On solving denominator →

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A hot metal bar is submerged in a large reservoir of water whose temperature is 60°F. The temperature of the bar 20 s after subm
True [87]

Answer:

A) cooling constant =  0.0101365

B) \frac{df}{dt} = k ( 60 - F )

c)  F(t) = 60 + 77.46e^{0.0101365t}

D)137.46 ⁰

Step-by-step explanation:

water temperature = 60⁰F

temperature of Bar after 20 seconds = 120⁰F

temperature of Bar after 60 seconds = 100⁰F

A) Determine the cooling constant K

The newton's law of cooling is given as

= \frac{df}{dt} = k(60 - F)

= ∫ \frac{df}{dt} = ∫ k(60 - F)

= ∫ \frac{df}{60 - F} = ∫ kdt

= In (60 -F) = -kt - c

       60 - F = e^{-kt-c}

      60 - F = C_{1} e^{-kt}       ( note : e^{-c} is a constant )

after 20 seconds

C_{1}e^{-k(20)} = 60 - 120 = -60  

therefore C_{1} = \frac{-60}{e^{-20k} } ------- equation 1

after 60 seconds

C_{1} e^{-k(60)} = 60 - 100 = - 40  

therefore C_{1} = \frac{-40}{e^{-60k} } -------- equation 2

solve equation 1 and equation 2 simultaneously

= \frac{-60}{e^{-20k} } = \frac{-40}{e^{-60k} }

= 6e^{20k} = 4e^{60k}

= \frac{6}{4} e^{40k} = In(6/4) = 40k

cooling constant (k) = In(6/4) / 40 = 0.40546 / 40 = 0.0101365

B) what is the differential equation  satisfied

substituting the value of k into the newtons law of cooling)

60 - F = C_{1} e^{0.0101365(t)}  

F(t) = 60 - C_{1} e^{0.0101365(t)}

The differential equation that the temperature F(t) of the bar

\frac{df}{dt} = k ( 60 - F )

C) The formula for F(t)

t = 20 , F = 120

F(t ) = 60 - C_{1} e^{0.0101365(t)}

120 = 60 - C_{1} e^{0.0101365(t)}

C_{1} e^{0.0101365(20)} = 60

C_{1} = 60 * 1.291 = 77.46

C1 = - 77.46⁰ as the temperature is decreasing

The formula for f(t)

= F(t) = 60 + 77.46e^{0.0101365t}

D) Temperature of the bar at the moment it is submerged

F(0) = 60 + 77.46e^{0.01013659(0)}

F(0) = 60 + 77.46(1)

     = 137.46⁰

5 0
4 years ago
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