Which is a more precise measurement 57.65lb or 34.9lb

Arrange the steps in the correct order to express cos3x in terms of cosx cos(2x)cos(x)-sin(2x)(x) [1-sin^2(x)]cos(x)-[2sin(x)cos

noname [10]

cos(3x) = cos(2x+x)

= cos(2x) cos (x) -sin(2x)sin(x)

$=(1-2sin^2(x)) cos(x)-(2 sin(x) cos(x)) sin(x)$

$= cos x -2sin^2(x) cos(x) -2sin^2(x) cos(x)$

$= cos(x) - 4 sin^2(x) cos(x)$

$cos(x)(1-4sin^2(x))$

$=cos(x(1-4(1-cos^2(x))$

$=cos(x)(1-4+4cos^2(x))$

$= cos(x)-4cos(x)+4cos^3(x)$

$=-3cos(x)+4cos^3(x)$

4 0

3 years ago

If a 50 ft flagpole casts a 95 ft shadow, then how long is the shadow that a 5 ft tall of man casts ?

lesantik [10]

9.5 feet,to find that you must find out how many times longer the shadow of the pole is than the pole, and to do that you must divide 95 by 50, in which you get 1.9. And then to find the length of the shadow the man casts you must multiply his height by 1.9.

8 0

3 years ago

HURRY PLEASE I DONT WANT TO FAIL

qwelly [4]

Answer: 62

Step-by-step explanation: In the red empty spot, the #5 would go there and in the blue spot the #6 would go there. 8x2=16 and 16+16= 32: t=This is for the to rectangle shapes on both sides. Now for the square in the middle, since the red spot is 5 and the blue is 6, 5x6= 30. So, 30+32=62.

4 0

2 years ago

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g A manufacturer is making cylindrical cans that hold 300 cm3. The dimensions of the can are not mandated, so to save manufactur

sdas [7]

Answer:

The dimensions that minimize the cost of materials for the cylinders have radii of about 3.628 cm and heights of about 7.256 cm.

Step-by-step explanation:

A cylindrical can holds 300 cubic centimeters, and we want to find the dimensions that minimize the cost for materials: that is, the dimensions that minimize the surface area.

Recall that the volume for a cylinder is given by:

$\displaystyle V = \pi r^2h$

Substitute:

$\displaystyle (300) = \pi r^2 h$

Solve for <em>h: </em>

$\displaystyle \frac{300}{\pi r^2} = h$

Recall that the surface area of a cylinder is given by:

$\displaystyle A = 2\pi r^2 + 2\pi rh$

We want to minimize this equation. To do so, we can find its critical points, since extrema (minima and maxima) occur at critical points.

First, substitute for <em>h</em>.

$\displaystyle \begin{aligned} A &= 2\pi r^2 + 2\pi r\left(\frac{300}{\pi r^2}\right) \\ \\ &=2\pi r^2 + \frac{600}{ r} \end{aligned}$

Find its derivative:

$\displaystyle A' = 4\pi r - \frac{600}{r^2}$

Solve for its zero(s):

$\displaystyle \begin{aligned} (0) &= 4\pi r - \frac{600}{r^2} \\ \\ 4\pi r - \frac{600}{r^2} &= 0 \\ \\ 4\pi r^3 - 600 &= 0 \\ \\ \pi r^3 &= 150 \\ \\ r &= \sqrt[3]{\frac{150}{\pi}} \approx 3.628\text{ cm}\end{aligned}$

Hence, the radius that minimizes the surface area will be about 3.628 centimeters.

Then the height will be:

$\displaystyle \begin{aligned} h&= \frac{300}{\pi\left( \sqrt[3]{\dfrac{150}{\pi}}\right)^2} \\ \\ &= \frac{60}{\pi \sqrt[3]{\dfrac{180}{\pi^2}}}\approx 7.25 6\text{ cm} \end{aligned}$