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kherson [118]
3 years ago
9

The goals against average (A) for a professional hockey goalie is determined using the formula A = 60 a equals 60 left-parenthes

is StartFraction g Over t EndFraction right-parenthesis.. In the formula, g represents the number of goals scored against the goalie and t represents the time played, in minutes. Which is an equivalent equation solved for g? StartFraction A t Over 60 EndFraction equals g.= g StartFraction A Over 60 t EndFraction equals g. = g StartFraction 60 A Over t EndFraction equals g. = g 60At = g
Mathematics
2 answers:
dangina [55]3 years ago
6 0

Answer:

Option 1 - \frac{At}{60}=g

It's A

Step-by-step explanation:

Marina CMI [18]3 years ago
4 0

Answer:

Option 1 - \frac{At}{60}=g

Step-by-step explanation:

Given : The goals against average (A) for a professional hockey goalie is determined using the formula A=60(\frac{g}{t} ). In the formula, g represents the number of goals scored against the goalie and t represents the time played, in minutes.

To find : Which is an equivalent equation solved for g?

Solution :

Solve the formula in terms of g,

A=60(\frac{g}{t})

Multiply both side by t,

At=60g

Divide both side by 60,

\frac{At}{60}=\frac{60g}{60}

\frac{At}{60}=g

Therefore, option 1 is correct.

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Answer:

55

Step-by-step explanation:

since, it is a parallelogram we can find the area easily just by multiplying the base and height

A=bh

A=55m2

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How many solutions does the following equation have? 4x+x+3=4x *
kvv77 [185]

Answer:

1 solution

Step-by-step explanation:

4x + x + 3 = 4x (Subtract 4x from both sides)

x + 3 = 0 (Subtract 3 from both sides)

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What is 5% of $40.00?
iogann1982 [59]

Answer:

$2

Step-by-step explanation:

40 x 0.05 = 2

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How to solve this kind of question?
S_A_V [24]
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5 0
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Read 2 more answers
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains liters of a dye solution with a
Alja [10]

Answer:

t = 460.52 min

Step-by-step explanation:

Here is the complete question

Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.

Solution

Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.

inflow = 0 (since the incoming water contains no dye)

outflow = concentration × rate of water inflow

Concentration = Quantity/volume = Q/200

outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.

So, Q' = inflow - outflow = 0 - Q/100

Q' = -Q/100 This is our differential equation. We solve it as follows

Q'/Q = -1/100

∫Q'/Q = ∫-1/100

㏑Q  = -t/100 + c

Q(t) = e^{(-t/100 + c)} = e^{(-t/100)}e^{c}  = Ae^{(-t/100)}\\Q(t) = Ae^{(-t/100)}

when t = 0, Q = 200 L × 1 g/L = 200 g

Q(0) = 200 = Ae^{(-0/100)} = Ae^{(0)} = A\\A = 200.\\So, Q(t) = 200e^{(-t/100)}

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

2 = 200e^{(-t/100)}\\\frac{2}{200} =  e^{(-t/100)}

㏑0.01 = -t/100

t = -100㏑0.01

t = 460.52 min

6 0
4 years ago
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