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3 years ago

If 0 < b ≤ 90 and cos(11b + 2) = sin(12b − 4), what is the value of b?

Mathematics

2 answers:

Elza [17]3 years ago

7 0

Answer:

b = 4

Step-by-step explanation:

The given expression is

Cos (11b + 2) = Sin (12b - 4)

Since, Sin(θ) = Cos(90 - θ)

Therefore, Cos (11b + 2) = Cos [90 - (12b - 4)]

Where 0 < b ≤ 90°

11b + 2 = 90 - (12b - 4)

11b + 2 = 94 - 12b

11b + 12b + 2 = 94 - 12b + 12b

11b + 12b + 2 - 2 = 94 - 2

23b = 92

$\frac{23b}{23}=\frac{92}{23}$

b = 4

Therefore, b = 4 will be the answer.

Elina [12.6K]3 years ago

4 0

Answer: B=4

Step-by-step explanation: Took the test for my geometry class

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Given f(x) = log(x+1), x >-1 and g(x) = x^2 + 2x, XER find (f•g)(1)

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Answer:

Like terms, functions may be combined by addition, subtraction, multiplication or division.

Example 1. Given f ( x ) = 2x + 1 and g ( x ) = x2

+ 2x – 1 find ( f + g ) ( x ) and

( f + g ) ( 2 )

Solution

Step 1. Find ( f + g ) ( x )

Since ( f + g ) ( x ) = f ( x ) + g ( x ) then;

( f + g ) ( x ) = ( 2x + 1 ) + (x2

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= 2x + 1 + x2

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= x

+ 4x

Step 2. Find ( f + g ) ( 2 )

To find the solution for ( f + g ) ( 2 ), evaluate the solution above for 2.

Since ( f + g ) ( x ) = x2

+ 4x then;

( f + g ) ( 2 ) = 22

+ 4(2)

= 4 + 8

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Example 2. Given f ( x ) = 2x – 5 and g ( x ) = 1 – x find ( f – g ) ( x ) and ( f – g ) ( 2 ).

Solution

Step 1. Find ( f – g ) ( x ).

( f – g ) ( x ) = f ( x ) – g ( x )

= ( 2x – 5 ) – ( 1 – x )

= 2x – 5 – 1 + x

= 3x – 6

Step 2. Find ( f – g ) ( 2 ).

( f – g ) ( x ) = 3x – 6

( f – g ) ( 2 ) = 3 (2) – 6

= 6 – 6

= 0

Example 3. Given f ( x ) = x2

+ 1 and g ( x ) = x – 4 , find ( f g ) ( x ) and ( f g ) ( 3 ).

Solution

Step 1. Solve for ( f g ) ( x ).

Since ( f g ) ( x ) = f ( x ) * g ( x ) , then

= (x2

+ 1 ) ( x – 4 )

= x

– 4 x2

+ x – 4 .

Step 2. Find ( f g ) ( 3 ).

Since ( f g ) ( x ) = x3

– 4 x2

+ x – 4, then

( f g ) ( 3 ) = (3)3

– 4 (3)2

+ (3) – 4

= 27 – 36 + 3 – 4

= -10

Example 4. Given f ( x ) = x + 1 and g ( x ) = x – 1 , find ( x ) and ( 3 ). f

⎛ ⎞ ⎜

⎝ ⎠

⎛ ⎞ ⎜

⎝ ⎠ ⎟ ⎟

Solution

Step 1. Solve for ( x ). f

⎛

⎜

⎝ ⎠

⎞

⎟

Since ( x ) = , then ( )

( )

f x

g x

⎛

⎜

⎝ ⎠

⎞

⎟

= ; x ≠ 1 1

−

Step 2 Find . ( ) 3 f

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

Since = , then 1

− ( ) f x

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

3 1

− ( ) 3 f

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

= 2

Step-by-step explanation:

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Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt

Leni [432]

Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = $\sqrt{Δx^2+Δy^2}$
d₁ = $\sqrt{(-2-1)^2+(1-4)^2}$
d₂ = $\sqrt{(x-1)^2+(y-4)^2}$
d₁ = d₂
∴ $\sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2}$ ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = $\pm \sqrt{9} = \pm 3$
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∴ y = 1 or y = 7

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3 years ago