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2 years ago

Let f(x)=cos(arctanx). What is the range of f?

Mathematics

1 answer:

Alenkasestr [34]2 years ago

4 0

-π/2 < arctan(x) < π/2

So cos(π/2) < cos(arctan(x)) < cos(0)

0 < cos(arctan(x)) < 1

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Whats the scientific notation of 0.258

loris [4]

3 x 10^-5.
Explanation: You move the decimal point 5 places to the right, hence it becomes 3 x 10 to the power of NEGATIVE 5 (moved 5 places right)

7 0

2 years ago

Please only answer if your certain it will mean a lot I can give extra points or brain list if you like or a thanks please don’t

BigorU [14]

Answer:

Ms. Cortez earns 500 dollars for working 30 hours. It's the last choice.

Step-by-step explanation:

You just have to substitute 30 into the equation in place of x, since x represents hours.

f(30) = 50 + 15(30)

f(30) = 50 + 450

f(30 = 500

3 0

2 years ago

The Kappa Iota Sigma Fraternity polled its members on the weekend party theme. The vote was as follows: six for toga, four for h

Nadusha1986 [10]

Answer:

note:

solution is attached

3 0

3 years ago

Which situation matches the inequality below?

mars1129 [50]

Answer:

C is the correct answer

8 0

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)

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

Dogs of the AMS.

4 0

2 years ago