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klio [65]
3 years ago
5

Find cos 0 for 0 = 30° The 0 has a weird line through it​

Mathematics
1 answer:
Nastasia [14]3 years ago
6 0

Answer:  \frac{\sqrt{3}}{2}

======================================================

Explanation:

The symbol \theta is the greek letter theta. It's often used in trigonometry for angles.

In this case, \theta = 30^{\circ}

So,

\cos(\theta) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2}

You'll need to use a reference sheet or the unit circle to determine the value of cos(30). Or you could use a 30-60-90 triangle template.

When using the unit circle, look in the upper right quadrant (which is quadrant 1). The angle 30 degrees is in this quadrant. Locate the terminal point and note how \frac{\sqrt{3}}{2} is the x coordinate of the terminal point. This is due to x = \cos(\theta)

Side note: \frac{\sqrt{3}}{2} \approx 0.866025

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You need to construct an open-top rectangular box with a square base that must hold a volume of exactly 475 cm3. The material fo
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Answer:

The dimensions of the box are:

x =  8,93 cm       and     h  =   5,95 cm

C(min) =  850,69 cents

Step-by-step explanation:

The volume of the box is:

V = x²*h          where    x is the side of the square base  and h the height

then    h  =  V/ x²  ⇒    h = 475 / x²

The total cost of box C is:

C  = C₁  +  4*C₂      Where C₁  and C₂  are the costs of the base and one lateral side respectevily

Then cost C =  8*x²   + 4* 6*h*x

The cost C as a function of x is

C(x)  =  8*x²  + (24* 475 /x² )*x

C(x)  =  8*x²  +  11400/x

Tacking derivatives on both sides of the equation

C´(x)  =  16*x -  11400/x²

C´(x)  =  0     ⇒    16*x  -  11400/x²  = 0

16*x³  =  11400     ⇒   x³  =  11400/16

x³ =  712,5

x  =  8,93  cm

and    h   =  475 / (8,93)²      ⇒      h  =  5,95  cm

C(min)  =  8*79,77  +  4* ( 8,93)*5,95

C(min)  =  638,16  +  212,53

C(min)  =  850,69 cents

To check if value x = 8,93 would make C(x) minimum we go to the second derivatives

C´´(x) =  16  +  22800/x³ > 0

Then we have a minimum of C at  x = 8,93

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