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Rina8888 [55]
3 years ago
6

Csc^3x-csc^2x-cscx+1=cot^2x(cscx-1)​

Mathematics
2 answers:
belka [17]3 years ago
5 0

Verify the identity:

csc^3x-csc^2x-cscx+1\qquad =cot^2x(cscx-1)\\\\csc^2x(cscx-1)-1(cscx-1) =\quad \downarrow\\\\(csc^2x-1)(cscx-1)\quad \qquad \ =\quad \downarrow\\\\\bigg(\dfrac{1}{sin^2x}-\dfrac{sin^2x}{sin^2x}\bigg)(cscx-1)\ =\quad \downarrow\\\\\\\bigg(\dfrac{1-sin^2x}{sin^2x}\bigg)(cscx-1)\qquad \ =\quad \downarrow\\\\\\\bigg(\dfrac{cos^2x}{sin^2x}\bigg)(cscx-1)\qquad \qquad =\quad \downarrow\\\\\\cot^2x(cscx-1)\qquad \qquad \qquad =cot^2x(cscx-1)\qquad \checkmark

Svetradugi [14.3K]3 years ago
5 0

Sorry, I messed up. So I changed it.

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The area of a rectangle is 70 m², and the length of the rectangle is 11 m less than three times the width. Find the dimensions o
11111nata11111 [884]

Step-by-step explanation:

the area of a rectangle is

length × width

in our case

length × width = 70 m²

length = 3×width - 11

we can use this second equation in the first and get

(3×width - 11) × width = 70

3×width² - 11×width - 70 = 0

the general solution to a quadratic equation is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 3

b = -11

c = -70

x = (11 ± sqrt((-11)² - 4×3×-70))/(2×3) =

= (11 ± sqrt(121 + 840))/6 =

= (11 ± sqrt(961))/6

x1 = (11 + 31)/6 = 42/6 = 7

x2 = (11 - 31)/6 = -20/6 = -10/3

the negative solution is not applicable for a length in an object.

so, only x1 = 7 m is our solution.

the width of the rectangle is 7 m.

the length = 3×width - 11 = 3×7 - 11 = 21-11 = 10 m

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2 years ago
Hurry plz What is the area of the given circle?
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Answer:

<em>600π feet</em>

Step-by-step explanation:

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The relationship between the actual air temperature x (in degrees Fahrenheit) and the temperature y adjust for wind chill (in de
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Step-by-step explanation:

the answer is in the above image

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icang [17]

Answer:

\int\limits {e^{2s} cos\frac{s}{4} ds    =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that  f(s) =  e^{2s} cos\frac{s}{4}

Now integrating

            \int\limits {f(s)} \, ds =  \int\limits {e^{2s} cos\frac{s}{4} ds

By using integration formula

   \int\limits { e^{ax} cos b x dx = \frac{e^{ax} }{a^{2}+b^{2}  } ( a cos b x + b sin b x )

<u><em>Step(ii):-</em></u>

 \int\limits {e^{2s} cos\frac{s}{4} ds    =   \frac{e^{2s} }{(2)^{2}+(\frac{1}{4}) ^{2}  } ( 2 cos (\frac{1}{4} ) s + \frac{1}{4}  sin \frac{1}{4}  s ))  

                    = \frac{e^{2s} }{(4+\frac{1}{16})} ( 2 cos (\frac{1}{4} ) s + \frac{1}{4}  sin \frac{1}{4}  s ))

                   = \frac{e^{2s} }{(\frac{65}{16} } ( \frac{8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s}{4}  ))

                 = 16 X\frac{e^{2s} }{65 } ( \frac{8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s}{4}  ))

                 =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

<u><em>Final answer:-</em></u>

\int\limits {e^{2s} cos\frac{s}{4} ds    =\frac{4 e^{2s} }{65 } ({8 cos (\frac{1}{4} ) s +  sin \frac{1}{4}  s} ))

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3 years ago
Solve the inequality and describe the solution set.<br><br> y – 6 ≥ 12
Marta_Voda [28]
Check image, hope it helps

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