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nikdorinn [45]
3 years ago
14

Given sinx+1/sinx=1+cscx, find a numerical value of one trigonometric function of x.

Mathematics
1 answer:
Nana76 [90]3 years ago
3 0

Answer:

x = π/2 or

x = π/2 + 2πn (for any value of n)

Step-by-step explanation:

We need to find the numerical value of the trigonometric function sinx+1/sinx=1+cscx

Subtract 1+cscx on both sides

sinx + 1/sinx -(1+csc x) = 1+cscx - (1+cscx)

sinx + 1/sinx -1 -csc x = 0

Taking LCM i.e, sinx and solving

(sinx)(sinx) + 1 - sinx -cscx(sinx) / sinx = 0

Multiplying sinx on both sides

sin^2 x + 1 -sinx - cscx(sinx) =0

as we know, cscx = 1/ sinx

sin^2 x + 1 -sinx - (1/sinx)(sinx) = 0

Solving,

sin^2 x + 1 -sinx - 1 = 0

sin^2 x - sin x =0

Let u = sinx

Putting it in above equation

u^2 - u =0

Solving the equation:

u= 1 , u= 0

Putting back the value of u

sin(x) = 1 and sin(x) = 0

x = sin ⁻¹ (1) and x = sin ⁻¹ (1) =0

x = 90° or π/2 and x = 0 (undefined for the question)

So, x = π/2 or

x = π/2 + 2πn (for any value of n)

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2x2-3 evaluate each expression. Show your work
nirvana33 [79]

Answer:

1

Step-by-step explanation:

2x2 = 4

4-3=1

5 0
3 years ago
Help me please ASAP
Oliga [24]

Answer:

30

Step-by-step explanation:

if im reading the qestuion right

so she had 80 bucks she spend 20 on a ticked so she has 60 left so if im reading it right she spent all of her money so 60 divde by 2 is 30.  

hoped this helped!

3 0
3 years ago
If x > 0 and y < 0, where is the point (x, y) located?
dlinn [17]

Answer:

Quadrant 4

Step-by-step explanation:

x > 0 means towards right of the origin

y < 0 means below the origin

So Quadrant 4

6 0
3 years ago
1/3 (x + 4 ) = 20 wt the answer
sashaice [31]

Answer:

x = 56

Step-by-step explanation:

To start, distribute the 1/3 onto the x and the 4 in the parenthesis. That leaves you with

1/3x + 4/3 = 20

We want the x value by itself so we're going to subtract the 4/3 from the left and the right side, leaving us with

1/3x + 4/3 = 20 - 4/3

1/3x = 18 & 2/3 (since we're dealing with fractions, I'm going to change the 18 & 2/3 into an improper fraction just to make the next step easier.)

1/3x = 56/3

The last step is to divide the 1/3 off of the x, so x is by itself. To divide a fraction by a fraction, just flip the fraction you're dividing with and multiply.

56/3 x 3/1 = 56 (56 x 3 is 168, and 3 x 1 is 3, so when you reduce 168/3, you get 56)

So, 56 is our answer, because after dividing off the 1/3, we're left with x = 56

3 0
3 years ago
La barbería El Caleño, tiene en promedio 120 clientes a la semana a
Luba_88 [7]

Queremos maximizar el precio de tal forma que los ingresos no disminuyan.

Ese maximo precio es: $14,040.6

Sabemos que actualmente el precio es:

p = $6,000

El número de clientes es:

C = 120

Actualmente los ingresos son el producto de esos dos números, es decir:

ingresos = $6,000*120 = $720,000

Ahora sabemos que por cada incremento de $700 en el precio, el número de clientes decrece en 10.

Entonces podemos escribir el número de clientes como una ecuación lineal.

C(p) = a*p + b

tal que tenemos dos puntos en esa linea:

($6,000, 120)

($6,700, 110)

La pendiente es:

a = \frac{110 - 120}{\$6,700 - \$6,000} = \frac{-10}{\$ 700}

Entonces tenemos:

C(p) = (-10/$700)*p + b

Sabemos que:

C($6,000) = 120 = (-10/$700)*$6,000 + b

                     120 = -85.71 + b

                     120 + 85.71 = b =

Entonces la ecuación lineal es:

C(p) = (-10/$700)*p + 205.71

Los ingresos serán dados por:

ingresos = C(p)*p = (-10/$700)*p^2 + 205.71*p

Y queremos maximizar p de tal forma que esto sea igual a lo que obtuvimos antes:

(-10/$700)*p^2 + 205.71*p = $720,000

Entonces debemos resolver la ecuación cuadratica:

(-10/$700)*p^2 + 205.71*p - $720,000 = 0.

Las soluciones son dadas por la formula de Bhaskara.

p = \frac{-205.71 \pm \sqrt{(205.71)^2 - 4*(-10/\$ 700)*\$ 720,000} }{2*(-10/\$ 700)} \\\\p = \frac{-205.71 \pm 195.45}{(-20/\$ 700)}

La solución de maximo valor es:

p = (-205.71 - 195.45)/(-20/$700) = $14,040.6

Sí quieres aprender más, puedes leer.

brainly.com/question/8926135

7 0
3 years ago
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