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aniked [119]
3 years ago
8

Find an exact value. Cos 75° negative square root of six plus square root of two quantity square root of six minus square root o

f two divided by four quantity negative square root of six plus square root of two divided by four square root of six minus square root of two
Mathematics
1 answer:
ehidna [41]3 years ago
6 0

Answer:

√2(√3 - 1)/4

Step-by-step explanation:

To find an exact value for Cos75°, we use the compound angle formula. Since 75° = 45° + 30°, Cos75° = Cos(45° + 30°).

Using Cos(A + B) = CosACosB - SinASinB where A = 45° and B = 30°,  

Cos75° = Cos(45° + 30°) = Cos45°Cos30° - Sin45°Sin30°

Now Cos45° = Sin45° = 1/√2 = √2/2, Cos30° = √3/2 and Sin30° = 1/2.

Substituting these values into the above equation, we have

Cos75° = Cos(45° + 30°)

= Cos45°Cos30° - Sin45°Sin30°

= √2/2 × √3/2 -  √2/2 × 1/2

= √6/4 -√2/4

= √2(√3 - 1)/4

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Tju [1.3M]

Answer:

21.68 minutes ≈ 21.7 minutes

Step-by-step explanation:

Given:

T=60+40e^{kt}

Initial temperature

T = 100°C

Final temperature = 60°C

Temperature after (t = 3 minutes) = 90°C

Now,

using the given equation

T=60+40e^{kt}

at T = 90°C and  t = 3 minutes

90=60+40e^{k(3)}

30=40e^{3k}

or

e^{3k}=\frac{3}{4}

taking the natural log both sides, we get

3k = \ln(\frac{3}{4})

or

3k = -0.2876

or

k = -0.09589

Therefore,

substituting k in 1 for time at temperature, T = 65°C

65=60+40e^{( -0.09589)t}

or

5=40e^{( -0.09589)t}

or

e^{( -0.09589)t}=\frac{5}{40}

or

e^{( -0.09589)t}=0.125

taking the natural log both the sides, we get

( -0.09589)t = ln(0.125)

or

( -0.09589)t = -2.0794

or

t = 21.68 minutes ≈ 21.7 minutes

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anzhelika [568]

Answer:

12

Step-by-step explanation:

The proportion is equal to \frac{x}{10}=\frac{42}{35}

We can simplify the right side by dividing by 7/7 to get the right side equal to 6/5. Now we can cross-multiply to get

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Dividing both sides by 5, we have the answer as

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Step-by-step explanation:

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Answer:

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Therefore the distance between the two on a number line is 3

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3 years ago
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Which ones do you need help on i’ll try to help you ???
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