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3 years ago

The temperature was -10 degrees. It warmed up 30 degrees. Then it got 40 degrees colder. What is the temperature now?

Mathematics

1 answer:

Murrr4er [49]3 years ago

3 0

Answer:

The temperature is -20 degrees.

Step-by-step explanation:

-10 + 30 = 20

20 - 40 = -20

Hope this helps!

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Solomon graphs a point on the coordinate plane. Which point did Solomon graph?

Mkey [24]

Answer:

Try your best! I know you can do it!

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4 years ago

At the beach, Nardia collected triple the number of seashells that Pierre did. Together, they collected 52 seashells. Which equa

MrRa [10]

Answer:

13 seashells

Step-by-step explanation:

p=number of seashells pierre collected

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3 years ago

Read 2 more answers

Ben consumes an energy drink that contains caffeine. After consuming the energy drink, the amount of caffeine in Ben's body decr

Airida [17]

Answer:

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

Step-by-step explanation:

After consuming the energy drink, the amount of caffeine in Ben's body decreases exponentially.

This means that the amount of caffeine after t hours is given by:

$A(t) = A(0)e^{-kt}$

In which A(0) is the initial amount and k is the decay rate, as a decimal.

The 10-hour decay factor for the number of mg of caffeine in Ben's body is 0.2722.

1 - 0.2722 = 0.7278, thus, $A(10) = 0.7278A(0)$ . We use this to find k.

$A(t) = A(0)e^{-kt}$

$0.7278A(0) = A(0)e^{-10k}$

$e^{-10k} = 0.7278$

$\ln{e^{-10k}} = \ln{0.7278}$

$-10k = \ln{0.7278}$

$k = -\frac{\ln{0.7278}}{10}$

$k = 0.03177289938
$

Then

$A(t) = A(0)e^{-0.03177289938t}$

What is the 5-hour growth/decay factor for the number of mg of caffeine in Ben's body?

We have to find find A(5), as a function of A(0). So

$A(5) = A(0)e^{-0.03177289938*5}$

$A(5) = 0.8531$

The decay factor is:

1 - 0.8531 = 0.1469

The 5-hour decay factor for the number of mg of caffeine in Ben's body is of 0.1469.

7 0

3 years ago

Evaluate. 3⋅23+52−18÷6 25 31 40 46

scoray [572]

Photo has the answer

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3 years ago

If a cup of coffee has temperature 97Â°C in a room where the ambient air temperature is 25Â°C, then, according to Newton's Law o

FinnZ [79.3K]

Answer:

The average temperature is $T_{a} = 81.95^oC$

Step-by-step explanation:

From the question we are told that

The temperature of the coffee after time t is $T(t) = 25 + 72 e^{[-\frac{t}{45} ]}$

Now the average temperature during the first 22 minutes i.e fro $0 \to 22$ minutes is mathematically evaluated as

$T_{a} = \frac{1}{22-0} \int\limits^{22}_{0} {25 +72 e^{[-\frac{t}{45} ]}} \, dx$

$T_{a} = \frac{1}{22} [25 t + 72 [\frac{e^{[-\frac{t}{45} ]}}{-\frac{1}{45} } ] ] \left| 22} \atop {0}} \right.$

$T_{a} = \frac{1}{22} [25 t - 3240e^{[-\frac{t}{45} ]} ] \left | 45} \atop {{0}} \right.$

$T_{a} = \frac{1}{22} [25 (22) - 3240e^{[-\frac{22}{45} ]} - (- 3240e^{0} )]$

$T_{a} = \frac{1}{22} [550 - 1987.12 + 3240]$

$T_{a} = 81.95^oC$

8 0

3 years ago