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Elina [12.6K]
4 years ago
10

Ryan drove for 4.5 hours to his uncle’s house. His uncle lives 234 miles away. Assuming Ryan drove at a constant rate, at what s

peed did he travel?
Mathematics
2 answers:
kari74 [83]4 years ago
6 0

Answer:52 i know this because i just did it on my work and got it right.

Step-by-step explanation:

Fantom [35]4 years ago
5 0
If you were to find the constant rate, you would have to take 4.5 and divide it into 234 to get the answer

You would get 52 as the constant speed

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Work out the answer to this problem to reduce to the simplest form 5/8-1/2=
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In order to subtract fractions, you need the denominators (bottom number) to be the same. \frac{1}{2} * \frac{4}{4} =  \frac{4}{8} . \frac{5}{8} - \frac{4}{8} =  \frac{1}{8}
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5 0
3 years ago
If point Q(3,-4) is rotated +270° about the origin, what are the coordinates of Q?
erica [24]

Answer:

Q' =(-4,-3)

Step-by-step explanation:

Given

Q = (3,-4)

Rotation = +270 degrees

Required

Determine the new coordinates of Q (i.e. Q')

When a point (x,y) is rotated +270° about the origin, the new coordinates is: (y,−x)

So:

Q = (3,-4)

Where:

x = 3 and y = -4

Q' will be:

Q' =(-4,-3)

8 0
3 years ago
Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g
AysviL [449]

Answer:

The quantity of salt at time t is m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} }), where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}

Where:

c - The salt concentration in the tank, as well at the exit of the tank, measured in \frac{pd}{gal}.

\frac{dc}{dt} - Concentration rate of change in the tank, measured in \frac{pd}{min}.

V - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

V \cdot \frac{dc}{dt} + 6\cdot c = 3

60\cdot \frac{dc}{dt}  + 6\cdot c = 3

\frac{dc}{dt} + \frac{1}{10}\cdot c = 3

This equation is solved as follows:

e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }

\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }

e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt

e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C

c = 30 + C\cdot e^{-\frac{t}{10} }

The initial concentration in the tank is:

c_{o} = \frac{10\,pd}{60\,gal}

c_{o} = 0.167\,\frac{pd}{gal}

Now, the integration constant is:

0.167 = 30 + C

C = -29.833

The solution of the differential equation is:

c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }

Now, the quantity of salt at time t is:

m_{salt} = V_{tank}\cdot c(t)

m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })

Where t is measured in minutes.

7 0
3 years ago
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