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3 years ago

9

Let ∠D be an acute angle such that cosD=0.33. Use a calculator to approximate the measure of ∠D to the nearest tenth of a degree

.

2 answers:

tia_tia [17]3 years ago

6 0

Answer:

.3

Step-by-step explanation:

.33 rounded equals .3

andreyandreev [35.5K]3 years ago

5 0

Answer:

PoInT tHrEe Is ThE aNsWeR

Step-by-step explanation:

.3

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Tresset [83]

Answer:

pretty sure its B

Step-by-step explanation:

5 0

3 years ago

Sharon wants to make a graph to show the relationship between the number of tickets sold and the price of tickets sold. She plot

Masteriza [31]

The statement that best describes the graph is “It will not be spread out across the entire coordinate plane because in Step 3 Sharon plotted the independent variable on the y-axis.”

To add, graphing is a powerful tool for representing and interpreting numerical data. In addition, graphs and knowledge of algebra can be used to find the relationship between variables plotted on a scatter chart.

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3 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.

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3 years ago

Evaluate the following double integral: xy dA D where the region D is the triangular region whose vertices are (0, 0), (0, 3), (

natulia [17]

Answer:

I= 84

Step-by-step explanation:

for

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy$

since D is the rectangle such that 0<x<3 , 0<y<3

$I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4$

4 0

2 years ago

Two professional baseball teams played a four game series. Attendance for the first three games was 126,503 people, what was the

ioda

Answer: 44,815 people; 126,503 + n = 171,318; n = 171,318 − 126,503

6 0

3 years ago