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GREYUIT [131]
2 years ago
14

For a ride on a rental scooter, Goran paid a $3 fee to start the scooter plus 11 cents per minute of the ride. The total bill fo

r Goran's ride was $23.79. For how many minutes did Goran ride the scooter?
Mathematics
1 answer:
lord [1]2 years ago
7 0

Answer:

Step-by-step explanation:

he paid a 3 dollar fee so we can remove that 23.79-3 = 20,79

now we devide 20,79 by 11 or 189 minutes.

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Find x=, if x-7=11 HURRY!!
Rashid [163]

Answer:

X = 18

Step-by-step explanation:

X - 7 = 11

Add 7 to both sides

X - 7 + 7 = 11+7

X = 18

6 0
3 years ago
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Please help its easy just i'm dvmb, just need this for my homework real soon!.. DON'T MIND THOSE ANSWERS SELECTED!!!!
trasher [3.6K]
<h3><em>Hey there today we will solve your problem,</em></h3>

Number 1

_______________

<h3>Definitions</h3>
  • Vertical angles - <em>either of two angles lying on opposite sides of two intersecting lines.</em>

<em />

  • Linear Pair - <em>A linear pair is a pair of adjacent angles formed when two lines intersect.</em>

_______________

Now we can solve Number 1, there is one pair of Vertical angles being ∠5 and ∠3

_______________

Number 2

Since we know the definition of a Linear Pair we can solve this problem also, the only Linear Pair that we can choose out the ones give to us is ∠4 and ∠3, because they are adjacent.

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{\huge{\boxed{\mathfrak{Answer}}}

  1. ∠5 and ∠3

    2. ∠4 and ∠3

8 0
2 years ago
Hey can someone help me?
RoseWind [281]
No sir .it’s 100 +18
4 0
3 years ago
The dimensions of a right rectangular prism are shown below.
sladkih [1.3K]

Answer:

216

Step-by-step explanation:

Im taking the topic assessment rn

5 0
3 years ago
The liquid base of an ice cream has an initial temperature of 86°C before it is placed in a freezer with a constant temperature
Karolina [17]

The temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

From Newton's law of cooling, we have that

T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt}

Where

(t) = \ time

T_{(t)} = \ the \ temperature \ of \ the \ body \ at \ time \ (t)

T_{s} = Surrounding \ temperature

T_{0} = Initial \ temperature \ of \ the \ body

k = constant

From the question,

T_{0} = 86 ^{o}C

T_{s} = -20 ^{o}C

∴ T_{0} - T_{s} = 86^{o}C - -20^{o}C = 86^{o}C +20^{o}C

T_{0} - T_{s} = 106^{o} C

Therefore, the equation T_{(t)}= T_{s}+(T_{0} - T_{s})e^{kt} becomes

T_{(t)}=-20+106 e^{kt}

Also, from the question

After 1 hour, the temperature of the ice-cream base has decreased to 58°C.

That is,

At time t = 1 \ hour, T_{(t)} = 58^{o}C

Then, we can write that

T_{(1)}=58 = -20+106 e^{k(1)}

Then, we get

58 = -20+106 e^{k(1)}

Now, solve for k

First collect like terms

58 +20 = 106 e^{k}

78 =106 e^{k}

Then,

e^{k} = \frac{78}{106}

e^{k} = 0.735849

Now, take the natural log of both sides

ln(e^{k}) =ln( 0.735849)

k = -0.30673

This is the value of the constant k

Now, for the temperature of the ice cream 2 hours after it was placed in the freezer, that is, at t = 2 \ hours

From

T_{(t)}=-20+106 e^{kt}

Then

T_{(2)}=-20+106 e^{(-0.30673 \times 2)}

T_{(2)}=-20+106 e^{-0.61346}

T_{(2)}=-20+106\times 0.5414741237

T_{(2)}=-20+57.396257

T_{(2)}=37.396257 \ ^{o}C

T_{(2)} \approxeq  37.40 \ ^{o}C

Hence, the temperature of the ice cream 2 hours after it was placed in the freezer is 37.40 °C

Learn more here: brainly.com/question/11689670

6 0
2 years ago
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