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c) Write an equivalent form of 1 / (1+sin(x)) without using fractions.

Multiply top/bottom by (1 - sin x) and we have

[1 - sinx] / [ 1 - sin^2x]  =

[1 - sin x ] / cos^2x =

1/cos^2x - sinx/ cos^2x =

sec^2x - tanx* secx =

secx [ secx - tan x]

There is a lot that can be factorised here but nothing cancels

  :(

Part a is not an identity.

If you try to establish:  sin(t) / ( 1 + cos(t) )   =   csc(t)

--->     sin(x) / ( 1 + cos(t) )  ·  [ ( 1 - cos(t) ) / ( 1 - cos(t) )             

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ ( 1 + cos(t) ) · ( 1 - cos(t) ) ]

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ 1 - cos²(t) ]                       since  1 - cos²(t) =  sin²(t) --->

--->     [ sin(x) · ( 1 - cos(t) ) ] / [ sin²(t)  ]       

--->     ( 1 - cos(t) ) / sin(t)

--->     1 / sin(t) - cos(t) / sin(t)

--->     csc(t) - cot(t)                                        not just csc(t) 

Part b:      ( sec²(t) - 1 ) / sec²(t)  =  sin²(t)

--->     ( sec²(t) - 1 ) / sec²(t)

--->     sec²(t) / sec²(t)  -  1 / sec²(t)

--->     1 - cos²(t)

--->     sin²(t)

Part c:     Write 1 / [ 1 + sin(x) ]  without fractions:

--->     1 / [ 1 + sin(x) ]  ·  [ ( 1 - sin(t) ) / ( 1 - sin(t) ) ]

--->     [ 1 · ( 1 - sin(t) ) ] / [ ( 1 + sin(t) ) · ( 1 - sin(t) ) ]

--->     [ 1 - sin(t) ] / [ 1 - sin²(t) ]

--->     [ 1 - sin(t) ] / [ cos²(t) ]

--->     1 / cos²(t)  -  sin(t) / cos²(t)

--->     sec²(t)  -  sin(t) / cos(t) · 1 / cos(t)

--->     sec²(t)  - tan(t) ·sec(t)

Hello Calpal!

Use trigonometric identities to show:

a) Sin(t) / (1+cos(t)) = csc(t)

b) (sec2(t)-1) / sec2(t) = sin2(t)

c) Write an equivalent form of 1 / (1+sin(x)) without using fractions

First off a)

a)

sin(t) / (1+cos(t)) = csc(t)

sin(t) / (1+cos(t)) = 1 / sin(t)

sin²(t) = 1 + cos(t)

1- cos²(t) = 1 + cos(t)

cos²(t) = - cos(t)

t ∈ {± periodically π/2; π; 3π/2}

Greeting asinus :- )  !

Hello fiora!

In the equation x^2+mx+n=0,m and n are intergers.

The only possible for x is -3.What is the value of m?

A.3

B.-3

C.6

D.-6

E.9

\(x^2+mx+n=0\)

\(x = -3\)

\(\left(x+ 3\right) ^{2}= x^{2} + 6x+ 9= 0\)

\(m = 6 \) 

\(Solution\ C\ is\ correct.\) 

Greeting asinus :- )  !

37=[√(x+167)]/2  multiply both sides by 2

74 = √(x+167)]   square  both sides

5476 = x + 167    subtract 167 from  both sides

5309 = x

I assume that this is

√ [16 ^(1/4 )] - 8 ^(1/3) + 9^(3/2)  =

√ [⁴√ 16 ] - ³√ 8  + [√9 ]³  =

√ 2  - 2  + 3³  =

√2 - 2 + 27  =

√2 + 25  =

25 + √2

approx.. 2.64926?

Hint :  the key to solving this kind of problem - if it can be solved - is to express the irrational part of the problem  as  2*sqrt(a * b)  and then finding two numbers, a and b, that sum to the integer part under the root

Also,  if the answer can be factored, the larger term in each pair of factors must come first...........

sqrt (49 - 20sqrt(6) ) =

sqrt (49  - 2sqrt (600) ) =

sqrt ( 49 - 2 sqrt (24 *25) ) =

sqrt 24 + 25 - 2sqrt (24 * 25) )  =

sqrt ( 24 - 2sqrt (24 * 25) + 25) =

sqrt (24 - 2sqrt (24)*sqrt(25) + 25)

And the expression inside the parentheses can be factored as :

[ sqrt (24) - sqrt (25) ] [ sqrt (24) - sqrt (25) ] =   [ the larger term must come first]

[sqrt (25) - sqrt(24)] [sqrt (25) - sqrt (24)]

[ sqrt (25) - sqrt (24)]^2

So

sqrt (49 - 20sqrt(6) ) =

sqrt ( [ sqrt (25) - sqrt (24)]^2 )  =

sqrt (25) - sqrt (24) =

5 - 2sqrt(6)

Notice that :

sqrt (14 + 8sqrt(3) )  =  sqrt ( 14 + sqrt(192)  ]  = sqrt (14 + 2sqrt(48)]  = sqrt (8 + 6 + 2 sqrt(8 * 6) =

sqrt ( 8 + 2sqrt (8 * 6) + 6)  =  8 + 2sqrt(8)sqrt(6) + 6

And the last expression can be factored as :

 [sqrt (8) + sqrt (6)]  * [ sqrt (8) + sqrt(6) ]  =

[sqrt (8) + sqrt6)]^2

So

sqrt (14 + 8sqrt(3) )  =

sqrt [ [sqrt (8) + sqrt6)]^2 ] =

sqrt (8) + sqrt (6)

Use the inclusion-exclusion principle to solve this problem.

|A or B| = |A| + |B| - |A and B|

Let set A = the 3s in [1...100] and set B = the 4s in [1...100]

|A| = 100 / 3 = 33

|B| = 100 / 4 = 25

|A and B| = 100 / (A*B) = 100 / 12 = 8

Therefore |A or B| = 33 + 25 - 8 = 50

The answer is 50.

(In this problem, || isn't the absolute value. It means the cardinality of the sets A and B.)

Problem:  x² - 5x + 6  =  2(x - 4)²  has answers  A  and  B.   Find the value of AB - A²

  x² - 5x + 6  =  2(x - 4)²  

 x² - 5x + 6  =  2(x - 4)(x - 4)

 x² - 5x + 6  =  2(x² - 8x + 16)

 x² - 5x + 6  =  2x² - 16x + 32

Subtract x² from both sides:     - 5x + 6  =  x² - 16x + 32

Add 5x to both sides:                          6  =  x² - 11x + 32

Subtract 6 frm both sides:                  0  =  x² - 11x + 26

Use the quadratic equation with a = 1, b = -11, and c = 26:

      A  =  [ -(-11) + sqrt( (11)² - 4(1)(26) ) ] / [ 2(1) ]           B  =  [ -(-11) - sqrt( (11)² - 4(1)(26) ) ] / [ 2(1) ]

      A  =  [11 + sqrt( 121 - 104 ) ] / [2]                                B  =  [11 + sqrt( 121 - 104 ) ] / [2]

      A  =  [ 11 + sqrt( 17 ) ] / [ 2 ]                                         B  =  [ 11 + sqrt( 17 ) ] / [ 2 ]

AB - A²      --->     [ 11 + sqrt( 17 ) ] / [ 2 ]  ·  [ 11 + sqrt( 17 ) ] / [ 2 ]   -   { [ 11 + sqrt( 17 ) ] / [ 2 ] }²

                 --->     [ 121 - 17 ] / 4     -     [ 121 + 11sqrt(17) + 11sqrt(17) + 17 ] / 4

                 --->          104 / 4      -     [ 138 + 22sqrt(17) ] / 4

                 --->     [ - 34 - 22sqrt(17) ] 4

14.99 * .084 ≈  $1.26

f (-2) = -2

So

 ( f  (f (-2) ) ) =

( f (-2) )  = -2

So

 ( f  (f (-2) ) )^-2  =

(-2)^-2  =

(1/-2)² =

1/4

The didstance traveled  after 9 complete bounces 

 15 + 30(.6) [ 1 - .6^9] / (1 - .6]  =  59.54 m = 59.5m

What is the sum of a finite geometric series whose first term is 5, common ratio is 3, and n^th term is 1215?

The n^th term of a geometric series can be found by using the formula:  a_n  =  a₁ · r^{n - 1}.

For this problem,  a_n = 1215,  a₁ = 5, and r = 3   --->   1215  =  5 · 3^{n - 1}

     divide both sides by 5:                                               243  =  3^{n - 1}

     take the log of both sides:                                   log(243)  =  log(3^{n - 1})

     simplify:                                                               log(243)  =  (n - 1) · log(3)

     divide both sides by log(3):                     log(243) / log(3)  =  n - 1

     simplify:                                                                           5  =  n - 1

     add 1 to both sides:                                                         n  =  6

To find the sum, use this formula:     Sum  =  a(1 - rⁿ) / (1 - r)

                                                          Sum  =  5(1 - 3⁶) / (1 - 3)

                                                          Sum  =  5(-728) / (-2)

                                                          Sum  =  -3640 / -2

                                                          Sum  =  1820 

2x^2 = kx-3k^2

2x^2 - kx + 3k^2  = 0

For this to have real roots, we must have that

k^2 - 4(2)(3k^2)  ≥ 0

k^2 - 24k^2  ≥ 0

-23k^2   ≥ 0

k^2  ≤ 0

This is only true when k = 0.........thus any other value of k will produce non-real solutions

HELP PLZ IM STUCK

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Are you Nauseated’s real life girlfriend? 

-2x2-4x+12 = 0  divide through by -2

x^2 + 2x - 6  = 0   using the quadratic formula where a = 1, b = 2 and c = -6,   x  = -1 ± √7

well to show work: 466646*65648754+487557664777456-8474785738444*8588885+948588666/74747+577-7=x 

so the answer is x= 6

Find a/b when 2log(a-2b)=log a + log b.

We have

2log (a - 2b)   =  log (a * b)

log (a - 2b)^2  = log (a * b)

Then......this implies that

(a - 2b)^2  = ab

a^2 - 4ab + 4b^2  =  ab

a^2 - 5ab + 4b^2 = 0

Factor

(a - 4b) (a - b)  = 0

The....setting both factors to 0 and solving

a =  4b  →  a/b  = 4

And 

a = b

a/b = 1

The first solution is good whenever a > 0

The second solution is good whenever a < 0

Could you tell me step by step? Thank you

40.25-27+46.5=239/4=59.75 as your final answer

Let   (4x - 16) / (3x - 4)   = p

So we have

p^2 + p  = 12

p^2 + p  - 12 = 0    factor

(p +4) (p - 3)  = 0      so p  = -4    or p =3

If p = -4, we have

(4x - 16) / (3x - 4)  = -4

4(x - 4) / (3x - 4)  = -4

(x - 4) / (3x - 4)  = -1

x - 4   = 4 - 3x

4x = 8

x = 2

If x = 3, we have

(4x - 16) / (3x - 4)  = 3

4x - 16  = 3(3x- 4)

4x - 16  = 9x - 12

- 4= 5x

x = -4/5

So......x = 2   is the largest value of x that solves the equation