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Ierofanga [76]
3 years ago
5

Cos(x + y) + cos(x − y) = 2 cos(x) cos(y)

Mathematics
2 answers:
Andre45 [30]3 years ago
7 0

Answer:

2cos(\frac{x+y+x-y}{2}) cos(\frac{x+y-x+y}{2})

2cos(\frac{2x}{2})cos(\frac{x+y-x+y}{2})

2cos(x)cos(\frac{2y}{2})

2cos(x)cos(y)

Alex17521 [72]3 years ago
6 0
   
\texttt{We use formula: }~~~  \boxed{\cos(x\pm y) = \cos (x) \cos (y) ~\mp ~\sin (x) \sin (y)}\\\\
\cos(x + y) + \cos(x - y) =\\\\
= \underbrace{\cos (x) \cos (y) -\sin (x) \sin (y)}_{\cos(x + y) } +\underbrace{\cos (x) \cos (y) +\sin (x) \sin (y)}_{\cos(x - y)} =\\\\
=\cos (x) \cos (y)+\cos (x) \cos (y)-\sin (x) \sin (y)+\sin (x) \sin (y)=\\\\
=\cos (x) \cos (y)(1+1) + \sin (x) \sin (y)(-1+1) = \\\\
=2\cos (x) \cos (y) + 0\sin (x) \sin (y) = \boxed{\boxed{2\cos (x) \cos (y)}}



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Write an equation of a line that passes through (-12, -14) with slope 6.
damaskus [11]

The slope intercept form of a line is y = mx + b

  • m = slope
  • b = y-intercept

Plug in the slope, 6, into m.

Rewrite the equation;

  • y = 6x + b
  • We need to find b, your y-intercept, to finish this equation.

Plug in your point coordinate, (x, y) ⇒ (-12, -14) into the equation.

  • -14 = 6(-12) + b

Solve for b to find the y-intercept.

  • -14 = -72 + b
  • 58 = b

Your new equation (your answer) is<em> </em>y = 6x + 58.

6 0
3 years ago
A painting is 24 in. Wide by 32in. Long. The width of a postcard reductions of the painting is 3in. How long is the postcard
s344n2d4d5 [400]

Answer:

4 inches.

Step-by-step explanation:

24 32

---- = -----

3 x

Cross multiply.

24x = (32)(3) = 96.

96 / 24 = 4.

Hope this helped and good luck! (:

7 0
3 years ago
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Sergeeva-Olga [200]

Answer:

Yes

Step-by-step explanation:

6 0
2 years ago
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Graph the inequalitly y&gt; 2/3x -1
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3 years ago
Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 162 square feet
Gnom [1K]

Answer:

The width and length of rectangle is 12.728 m

Step-by-step explanation:

Let the length of the rectangle = L

let the width of the rectangle = W

The subjective function is given by;

F(p) = 2(L + W)

F = 2L + 2W

Area of the rectangle is given by;

A = LW

LW = 162 ft²

L = 162 / W

Substitute in the value of L into subjective function;

f = 2l + 2w\\\\f = 2(\frac{162}{w} )+2w\\\\f = \frac{324}{w} + 2w\\\\\frac{df}{dw} = \frac{-324}{w^2} +2\\\\

Take the second derivative of the function, to check if it will given a minimum perimeter

\frac{d^2f}{dw^2}= \frac{648}{w^3} \\\\Thus, \frac{d^2f}{dw^2}>0, \ since,\frac{648}{w^3} >0 \ (minimum \ function \ verified)

Determine the critical points of the first derivative;

df/dw = 0

\frac{-324}{w^2} +2 = 0\\\\-324 + 2w^2=0\\\\2w^2 = 324\\\\w^2 = \frac{324}{2} \\\\w^2 = 162\\\\w= \sqrt{162}\\\\w = 12.728 \ m

L = 162 / 12.728

L = 12.728 m

Therefore, the width and length of rectangle is 12.728 m

3 0
3 years ago
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