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3 years ago

Given: cos a = -3/5, sin b = 5/13, and a and b are second-quadrant angles; find sin(a - b).

Mathematics

1 answer:

Vladimir79 [104]3 years ago

7 0

Answer:

63/65

Step-by-step explanation:

Recall that sin(a-b) = sin(a)cos(b) - cos(a)sin(b)

Therefore, we have:

sin(a-b) = sin(a)cos(b) - (-3/5)(5/13)

sin(a-b) = sin(a)cos(b) + 3/13

Since sin(x) = opposite/hypotenuse and cos(x) = adjacent/hypotenuse, then if cos(a) = -3/5, sin(a) = 4/5. Likewise, if sin(b) = 5/13, then cos(b) = 12/13.

Therefore, sin(a-b) = (4/5)(12/13) + 3/13 = 48/65 + 3/13 = 48/65 + 15/65 = 63/65

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A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and

nasty-shy [4]

Answer:

correct option is C) 2.8

Step-by-step explanation:

given data

string vibrates form = 8 loops

in water loop formed = 10 loops

solution

we consider mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = $\rho$

case (1)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

so here

$l = \frac{8 \lambda _1}{2}$ ..............1

$l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}$

and we know velocity is express as

velocity = frequency × wavelength .....................2

$\sqrt{\frac{Tension}{mass\ per\ unit \length }}$ = f × $\lambda_1$

here tension = mg

$\sqrt{\frac{mg}{\mu}}$ = f × $\lambda_1$ ..........................3

and

case (2)

when 8 loop form with 2 adjacent node is $\frac{\lambda }{2}$

$l = \frac{10 \lambda _1}{2}$ ..............4

$l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}$

when block is immersed

equilibrium eq will be

Tenion + force of buoyancy = mg

T + v × $\rho$ × g = mg

and

T = v × $\rho$ - v × $\rho$ × g

from equation 2

f × $\lambda_2$ = f × $\frac{1}{5}$

$\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}$ .......................5

now we divide eq 5 by the eq 3

$\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}$

solve irt we get

$1 - \frac{\rho _{stone}}{\rho _{water}} = \frac{16}{25}$

relative density $\frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}$

relative density = 2.78 ≈ 2.8

so correct option is C) 2.8

3 0

4 years ago

Solve the following system of equations by graphing.

never [62]

Answer:

(x,y)=(3,1)

Step-by-step explanation:

8 0

3 years ago

A total of 243 tickets were sold for the school play. They were either adult tickets or student tickets. The number of student t

alina1380 [7]

Answer:

81 adult tickets

Step-by-step explanation:

Step 1:

Set up a system of equations

Since the adult tickets and student tickets add to 243, we can do

$a+s=243$

and the student tickets sold were 2x the adult tickets, we can do

$s=2a$

We can replace s in the first equation adding to 243 with 2a

$a+s(2a)=243\\3a=243\\$

Step 2:

Solve for a. We can divide both sides by 3.

$a=81$

Therefore, there were 81 adult tickets sold.

Step 3:

We can check our work! Since we learned earlier the student tickets are 2x adult tickets, we can multiply 81 by 2 and add that to 81 to see if it equals 243!

$s=2a\\s=2(81)\\s=162$