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Liono4ka [1.6K]

2 years ago

14

The line of reflection is the ____. y-axis, center of rotation, x-axis

1 answer:

nydimaria [60]2 years ago

8 0

Answer:The line of reflection is the y axis

Step-by-step explanation:

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Subtract <br><br> \frac{3n+2}{n-4}-\frac{n-6}{n+4}

aalyn [17]

(3n+2)/(n-4) - (n-6)/(n+4)

common denominator (n-4)(n+4)

{(n+4)(3n+2)-(n-4)(n-6)}/{(n-4)(n+4)}

Use the foil method:

{(3n²+14n+8)-(n²-10n+24)}/{(n-4)(n+4)}

distribute negative sign:

{(3n²+14n+8-n²+10n-24)}/{(n-4)(n+4)}

subtract:

(2n²+24n-16)/{(n-4)(n+4)}

take out 2:

2{n²+12n-8}/{(n-4)(n+4)}

4 0

3 years ago

Read 2 more answers

Ahren’s birthday, a, is less than 6 days away. Which inequality represents Ahran’s birthday? PLz help me

Svetradugi [14.3K]

Answer:

the letter " a " because uts just making the sentence a run on

4 0

3 years ago

Find the area.<br> 11 cm<br> 8 cm<br> 7 cm

HACTEHA [7]

66.5cm

Split the shape into a square and triangle and find the area separately

Square: 8*7=57
Triangle: 7*3/2 = 10.5
=66.5

7 0

2 years ago

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Solve (-5/2)x ‐ 5 = -55

Dmitry_Shevchenko [17]

Answer: 5

Step-by-step explanation: add 5 on each side and you get -50 then divide -5/2 with -50 and get 5

5 0

2 years ago

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

so

$\sqrt{\frac{mg}{\mu}}$ = 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

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

so

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

3 years ago