Enter the value that makes the given equation true. Y+396.23 =574.54

The complement of an acute angle is an obtuse angle. <br><br> A.) True<br><br> B.) False

PilotLPTM [1.2K]

Your answer is False because obtuse angles are wide acute angles are right in the like this:^ but turn it to the west a right angle is like an L so your answer is False.

Your answer is False or B.

8 0

3 years ago

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If a hiker starts at sea level and increases her elevation at a rate of 1 foot per minute what will her elevation be in 20 minut

sattari [20]

Answer:

20 feet above sea level.

Explanation:

This can be solved by looking at the Rate of Change. It is given that for ONE minute that passes by, her elevation increases by ONE foot.

With this rate, after 20 minutes, the elevation can be shown as:

20 min × 1 foot= 20 feet.

4 0

3 years ago

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Organization is the key to an effective speech. Proper organization allows for a logical flow of words and ideas in a speech, ma

GREYUIT [131]

Answer:

This seems more like an answer then a question, did you perhaps mean to put this on someone elses question?

Step-by-step explanation:

4 0

3 years ago

(-3x^2+4x-2)/(-3x+19)

prohojiy [21]

\left[x _{1}\right] = \left[ \frac{2}{3}+\left( \frac{-1}{3}\,i \right) \,\sqrt{2}\right][x1]=[32+(3−1i)√2] totally answer

7 0

3 years ago

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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

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