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3 years ago

5

A right rectangular pyramid is sliced parallel to the base, as shown.

1 answer:

Natali5045456 [20]3 years ago

6 0

#1 15cm^2. ====> Good Luck! <====

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Factor 12y^2 + 5y - 2 completely.

guapka [62]

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2 years ago

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Before leaving to visit Mexico, levant traded 270 american dollars and received 3000 mexican pesos. When he returned from mexico

Makovka662 [10]

Levant will receive about 9 american dollars when he exchanges his pesos.

According to this question, 1 american dollar is worth about 11 pesos. In order to find that, you need to divide 3000/270 which gives you around 11. In order to find out how many dollars he will receive for his pesos, you need to divide 100 by 11 which gives you about 9 dollars.

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What is the decimal equivalent of the fraction?

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3 years ago

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(2x^2-5x+1)+(x^2-x-4)

maksim [4K]

Answer:

3x^2-6x -3

Step-by-step explanation:

(2x^2-5x+1)+(x^2-x-4)

we sum all coefficient according to the exponent of the variable

2x^2-5x+1+x^2-x-4 = 3x^2-6x -3

5 0

3 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