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xxTIMURxx [149]
3 years ago
8

Evaluate 2 tan 45 - tan 60​

Mathematics
2 answers:
Darya [45]3 years ago
6 0

Answer:

2 - √3

Step-by-step explanation:

tan45 = 1

tan60 = √3

2 tan 45 - tan 60​

2(1) - √3

=> 2 - √3

hoa [83]3 years ago
4 0

Answer:

In degrees: 0.27 (rounded)

In radians: 2.92 (rounded)

Step-by-step explanation:

<em>Brainliest, please!</em>

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leonid [27]

Answer:

B) (1/2, -8)

Step-by-step explanation:

(1, -6) and (0, -10)

Midpoint formula:

((x1+x2)/2, (y1+y2)/2)

Solving for x:

(x1+x2)/2

(1 + 0)/2

1/2

Solving for y:

(y1+y2)/2

(-6-10)/2

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2 years ago
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Answer:

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Step-by-step explanation:

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3 years ago
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gulaghasi [49]

Answer:yes

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2 years ago
Use &lt; , &gt; , = , ≤ , ≥ to complete these statements. 1) if x &lt; y , then y___ x, 2) if x &gt; 5 and 5 &gt; y, then x ___
vodomira [7]

Answer:

1) y > x

2) x > 5 > y

3) x ≤ 4  ( or 4 ≥ x )

4) x = y

Step-by-step explanation:

1) 1. y is greater than x

  2. so y > x

2) 1. 5 is less than x and more than y

   2. we can write this like x > 5 > y

   3. you can just squish them together as a shortcut usually: x > 5 and 5 > y      -->      x > 5 > y

3) 1. x is for at most, we can say this as x is less than or equal to 4

   2. this is written as x ≤ 4  or  4 ≥ x

4) 1. x - y = 0 so x and y must be the same value

   2. therefore x = y

4 0
3 years ago
The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),
pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}

(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}

When t=0, S(t)=15

15=30+Ce^{-k*0}\\C=15-30\\C=-15

When t = 2, S(t)=20

20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}

Therefore:

S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22

8 0
3 years ago
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