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3 years ago

Part B is easy I just need part A

Mathematics

1 answer:

lions [1.4K]3 years ago

3 0

Answer:

80 feet

Step-by-step explanation:

According to the tangent-secant theorem, the square of the tangent length is equal to the product of the secant segment and the external portion.

In other words:

30² = (10 + x) (10)

900 = 100 + 10x

800 = 10x

x = 80

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Let X be a positive continuous random variable with density fXpxq. Let Y " lnpXq. (a) Find the density fY pyq of Y in terms of t

salantis [7]

Answer:

X density = fXpxq and

Y" =InpXq

Now to find Y density FYpyq interms of the density of X we compare the density of X with Y"

fX = In

And PXq =pxq

Thus replacing x with y,

PXq = pyq

(a) Hence the density of Y is FYpyq

(b) at p0, fYpyq =fYp0q= 0

At 5s, FYpyq =5

7 0

3 years ago

Read 2 more answers

Translate this phrase into an algebraic expression.

Roman55 [17]

Answer:

8+2j like you say let j be the variable and more than means+

7 0

3 years ago

Read 2 more answers

Anyone know what 68010+01410 is

alex41 [277]

Put the first number on top then start adding 6+0, 8+1, 0+4, 1+1, 0+0 and the answer is 69420

7 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

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

3 years ago

Solve the equation. Round the answer to the nearest tenth.<br> 5*6^3n=20

solniwko [45]

Step-by-step explanation:

5×6'3n=20

30'3n=20

3n=20-30

3n/3= -10/3

n= -3

7 0

3 years ago