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Paladinen [302]
2 years ago
8

You and your family go out to dinner to build a total of $54 plan to leave 20% tip how much tip should you leave

Mathematics
1 answer:
loris [4]2 years ago
3 0

Answer:

$10.8 is left.

Step-by-step explanation:

Given that,

Total amount = $54

Left amount = 20%

We need to find how much tip should you leave. It means we need to find 20% of 54. So,

A=\dfrac{20}{100}\times 54\\\\=\$10.8

So, $10.8 is left.

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Step-by-step explanation:

the ratio is 3 :1

here is your answer

6 0
3 years ago
- The temperature of a hot liquid is 100 degrees. The liquid is placed in a refrigerator
oee [108]

Answer:

PLZ GIVE BRAINLESISET

Calculus Newton's Law of Cooling

Sources: #1 – 5: Smith/Minton Calculus 4th ed. #6 – 8: Thomas/Finney Calculus 9th ed.

Use Newton's Law of Cooling T −TS = (TO −TS )e−kt

( ) to solve the following. Round temperature

answers to the nearest tenth of a degree, and time (duration) answers to the nearest hundredth of a

minute.

1) A cup of fast-food coffee is 180°F when freshly poured. After 2 minutes in a room at 70°F, the coffee has

cooled to 165°F. Find the time that it will take for the coffee to cool to 120°F.

165− 70 = (180 − 70)e−2 k

95 =110e−2 k

e−2 k = 95

110

k =

ln

95

110

"

#

$ %

&

'

−2

120 − 70 = (180 − 70)e

−

ln 95

110

"

#

$ %

&

'

−2

"

#

$

$

$

$

%

&

'

'

'

'

t

50 =110e

ln 95

110

!

"

# $

%

&

2

!

"

#

#

#

#

$

%

&

&

&

&

t

e

ln 95

110

!

"

# $

%

&

2

!

"

#

#

#

#

$

%

&

&

&

&

t

= 50

110

t =

ln

50

110

!

"

# $

%

&

ln

95

110

!

"

# $

%

&

2

!

"

#

#

#

#

$

%

&

&

&

&

≈10.76 minutes

2) A bowl of porridge at 200°F (too hot) is placed in a 70°F room. One minute later the porridge has cooled to

180°F. When will the temperature be 120°F (just right)?

180 − 70 = (200 − 70)e−k

110 =130e−k

e−k = 11

13

k = −ln

11

13

"

#

$ %

&

'

120 − 70 = (200 − 70)e

ln 11

13

!

"

# $

%

&t

50 =130e

ln 11

13

!

"

# $

%

&t

e

ln 11

13

!

"

# $

%

&t

= 50

130

t =

ln

5

13

!

"

# $

%

&

ln

11

13

!

"

# $

%

&

≈ 5.72 minutes

3) A smaller bowl of porridge served at 200°F cools to 160°F in one minute. What temperature (too cold) will

this porridge be when the bowl of exercise 2 has reached 120°F (just right)?

160 − 70 = (200 − 70)e−k

90 =130e−k

e−k = 9

13

k = −ln

9

13

"

#

$ %

&

'

T − 70 = (200 − 70)e

ln 9

13

"

#

$ %

&

'(5.72)

T =130e

ln 9

13

"

#

$ %

&

'(5,72)

+ 70

T ≈ 85.9°

4) A cold drink is poured out at 50°F. After 2 minutes of sitting in a 70°F room, its temperature has risen to

56°F.

A) What will the temperature be after 10 minutes?

B) When will the drink have warmed to 66°F?

A)

56 − 70 = (50 − 70)e−2 k

−14 = −20e−2 k

e−2 k = 7

10

k =

ln

7

10

"

#

$ %

&

'

−2

T − 70 = (50 − 70)e

ln 7

10

!

"

# $

%

&

2 (10)

T = 70 +(−20)e

ln 7

10

!

"

# $

%

&

2 (10)

≈ 66.6°F

B)

66 − 70 = (50 − 70)e

ln(0.7)

2 t

−4 = −20e

ln(0.7)

2 t

e

ln(0.7)

2 t

= 4

20

t = ln(0.2)

ln(0.7)

2

≈ 9.02 minutes after it is poured.

5) At 10:07 pm, you find a secret agent murdered. Next to him is a martini that got shaken before he could stir

it. Room temperature is 70°F. The martini warms from 60°F to 61°F in the 2 minutes from 10:07 pm to

10:09 pm. If the secret agent's martinis are always served at 40°F, what was the time of death (rounded to the

nearest minute)?

61− 70 = (60 − 70)e−2 k

−9 = −10e−2 k

e−2 k = 9

10

k = ln(0.9)

−2

60 − 70 = (40 − 70)e

ln(0.9)

2 t

−10 = −30e

ln(0.9)

2 t

e

ln(0.9)

2 t

= 1

3

t =

ln

1

3

!

"

# $

%

&

ln(0.9)

2

≈ 20.85 minutes

The agent was murdered

at approx. 9:46 pm.

6) A hard-boiled egg at 98°C is put into a sink of 18°C water. After 5 minutes, the egg's temperature is 38°C.

Assuming that the surrounding water has not warmed appreciably, how much longer will it take the egg to

reach 20°C?

38−18 = (98−18)e−5k

20 = 80e−5k

e−5k = 20

80

k = ln(0.25)

−5

20 −18 = (38−18)e

ln(0.25)

5 t

2 = 20e

ln(0.25)

5 t

e

ln(0.25)

5 t

= 2

20

t = ln(0.1)

ln(0.25)

5

≈ 8.30 minutes

7) Suppose that a cup of soup cooled from 90°C to 60°C after 10 minutes in a room whose temperature

was 20°C.

A) How much longer would it take the soup to cool to 35°C?

60 − 20 = (90 − 20)e−10 k

40 = 70e−10 k

e−10 k = 40

70

k =

ln

4

7

!

"

# $

%

&

−10

35− 20 = (90 − 20)e

ln 4

7

!

"

# $

%

&

10 t

15 = 70e

ln 4

7

!

"

# $

%

&

10 t

e

ln 4

7

!

"

# $

%

&

10 t

= 15

70

t =

ln

3

14

!

"

# $

%

&

ln

4

7

!

"

# $

%

&

10

≈ 27.53 minutes

B) Instead of being left to stand in the room, the cup of 90°C soup is placed in a freezer whose temperature

is -15°C, and it took 5 minutes to cool to 60°C. How long will it take the soup to cool from 90°C

to 35°C?

60 −(−15) = (90 −(−15))e−5k

75 =105e−5k

e−5k = 75

105

k =

ln

5

7

!

"

# $

%

&

−5

35−(−15) = (90 −(−15))e

ln 5

7

!

"

# $

%

&

5 t

50 =105e

ln 5

7

!

"

# $

%

&

5 t

e

ln 5

7

!

"

# $

%

&

5 t

= 50

105

t =

ln

10

21

!

"

# $

%

&

ln

5

7

!

"

# $

%

&

5

≈11.03 minutes

8) A pan of warm water (46°C) was put into a refrigerator. Ten minutes later, the water's temperature was 39°C.

10 minutes after that it was 33°C. Use Newton's Law of Cooling to estimate the temperature of the

refrigerator.

39 −TS = (46 −TS )e−10 k ⇒ e−10 k = 39 −TS

46 −TS

33−TS = (39 −TS )e−10 k ⇒ e−10 k = 33−TS

39 −TS

39 −TS

46 −TS

= 33−TS

39 −TS

(39 −TS )

2

= (33−TS )(46 −TS )

1521− 78TS +TS

2 =1518− 79TS +TS

2

TS = −3°C

5 0
3 years ago
Which values are solutions to the inequality below?
s344n2d4d5 [400]

Answer:

A

B

E

Step-by-step explanation:

If you find what 11 squared is: 121

All number greater than or equal to 121 will be solutions.

3 0
2 years ago
Write 99÷12 as a decimal
gladu [14]

Answer:

8.25

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
What are the features of the parabola modeled by the equation f(x) = x2 - 114x ?
notsponge [240]

Answer:

Answers are in bold type

Step-by-step explanation:

f(x) = x^{2} -144x

The parabola opens up, so has a minimum at the vertex.

Let (h, k) be the vertex

h = -b/2a = - (-144)/2(1) = 57

k = 57^2 - 144(57) = 3249 - 6498 = -3249

Therefore, the vertex is (57, -3249)

The minimum value is -3249

The domain is the set of real numbers.  

The  range = {y | y ≥ -3249}

The function decreases when -∞ < x < 57  and increases when 57 > x > ∞

The x - intercepts:  x^{2} -144x = 0

                                x(x - 114x) = 0

                                x = 0 or x = 114

x-intercepts are (0, 0) and (0, 114)

When x = 0, then we get the y-intercept.  So, 0^2 - 114(0) = 0

y-intercept is (0, 0)

3 0
3 years ago
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