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3 years ago

What is the domain of the function represented by these ordered pairs ?{(-2,1), (0,0), (3,-1), (-1,7), (5,7)}

Mathematics

1 answer:

Mekhanik [1.2K]3 years ago

4 0

Answer:

{-2,-1,0,3,5}

Step-by-step explanation:

Domain = all the x-values in the set

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ehidna [41]

Answer:

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3 years ago

Write the word sentence as an equation.

Katyanochek1 [597]

The answer you are looking for is -5

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3 years ago

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Han recorded the number of pages that he read each day for five days. the dot plot shows his data. pages the mean, or average, f

Paul [167]

If the mean of the data set is 32.6. Then the number of pages he read in 5 days is 163.

Mean is simply defined as the average of the given set of numbers. The mean is considered as one of the measures of central tendencies in statistics. The mean is said to be an arithmetic mean. It is the ratio of the sum of the observation to the total number of observations.

Han recorded the number of pages that he read each day for five days.

If the mean of the data set is 32.6 pages. Then the total number of the pages he read till now, will be

Let the total number of the pages be x. Then we have

$\rm \dfrac{x}{5} = 32.6\\\\x \ = 5 \times 32.6 \\\\x \ = 163$

More about the mean link is given below.

brainly.com/question/521501

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2 years ago

A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where the

Vedmedyk [2.9K]

(a) Using Newton's Law of Cooling, $\dfrac{dT}{dt} = k(T - T_s)$ , we have $\dfrac{dT}{dt} = k(T - 75)$ where T is temperature after T minutes.
Separate by dividing both sides by T - 75 to get $\dfrac{dT}{T - 75} = k dt$ . Integrate both sides to get $\ln|T - 75| = kt + C$ .

Since $T(0) = 185$ , we solve for C:
$|185 - 75| = k(0) + C\ \Rightarrow\ C = \ln 110$
So we get $\ln|T - 75| = kt + \ln 110$ . Use T(30) = 150 to solve for k:
$\ln| 150 - 75 | = 30k + \ln 110\ \Rightarrow\ \ln 75 - \ln 110= 30k \Rightarrow \\ k= \frac{1}{30}\ln (75/110) = \frac{1}{30}\ln(15/22)$

So

$\ln|T - 75| = kt + \ln 110 \Rightarrow |T - 75| = e^{kt + \ln110} \Rightarrow \\ \\
|T - 75| = 110e^{kt} \Rightarrow T - 75 = \pm110e^{(1/30)\ln(15/22)t} \Rightarrow \\
T = 75 \pm110e^{(1/30)\ln(15/22)t}$

But choose Positive because T > 75. Temp of turkey can't go under.

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} \\
T(45) = 75 + 110e^{(1/30)\ln(15/22)(45)} = 136.929 \approx 137{}^{\circ}F$

(b)

$T(t) = 75 + 110e^{(1/30)\ln(15/22)t} = 75 + 110(15/22)^{t/30} \\
100 = 75 + 110(15/22)^{t/30} \\
25 = 110(15/22)^{t/30} 
\frac{25}{110} = (15/22)^{t/30} \\
\ln(25/110) / ln(15/22) = t/30 \\
t = 30\ln(25/110) / ln(15/22) \approx 116\ \mathrm{min}$

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3 years ago

Help please!! :)))) Thanks!

baherus [9]

The answer is C. 16!!

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3 years ago