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Dafna11 [192]
2 years ago
8

To rent a certain meeting room, a college charges a reservation fee of $ 46 and an additional fee of $ 6.40 per hour. The chemis

try club wants to spend at most $ 78.00 on renting the meeting room. What are the possible amounts of time for which they could rent the meeting room? Use t for the number of hours the meeting room is rented, and solve your inequality for t .
Mathematics
1 answer:
lisabon 2012 [21]2 years ago
8 0

Answer:

Up to 5 hours

Step-by-step explanation:

<u>Given:</u>

  • One off fee = $46
  • Hourly rent = $6.4 / hr
  • Amount limitation= $78

<u>Equation to reflect the condition:</u>

  • 78 = 46 + 6.4t
  • 6.4t = 78 - 46
  • 6.4t = 32
  • t = 32/6.4
  • t = 5 hours max

So, the meeting room can be rented for up to 5 hours with $78

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map of a city uses the scale 1 cm equals 50 m. On the​ map, South Street is 25 cm long. If there is a traffic cone at the start
aivan3 [116]

Answer:

125 traffic cones

Step-by-step explanation:

In order to find the total number of meters, we can set up a proportion to find based on the ration of cm to m:

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3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
Read 2 more answers
Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt
ser-zykov [4K]

Answer:

I(t)=\frac{1}{3}(1-e^{-30t})

Step-by-step explanation:

We are given that

\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}

R=150 ohm

L=5 H

V(t)=10 V

P=\frac{R}{L}=\frac{150}{5}=30

I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}

I(t)\times I.F=\int e^{30 t}\times 10 dt+C

I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C

I(t)=\frac{1}{3}+Ce^{-30 t}

I(0)=0

Substitute t=0

0=\frac{1}{3}+C

C=-\frac{1}{3}

Substitute the values

I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}

I(t)=\frac{1}{3}(1-e^{-30t})

5 0
3 years ago
2) Eight is 5 less than<br> twice the number R.
posledela

Answer:

8-5 times R

Step-by-step explanation:

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7 0
3 years ago
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