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4 years ago

Solve for k. 5 + 2k = 7

Mathematics

2 answers:

ryzh [129]4 years ago

7 0

So 5 + 2k = 7k

Hope this helps

igomit [66]4 years ago

6 0

Answer:

k = 1

Step-by-step explanation:

5 + 2k = 7

2k = 7 - 5

k = 2/2

k = 1

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Using the midpoint method (show your work), calculate the price elasticity of demand when the price of a barrel of gosum berries

Setler79 [48]

Answer:

0.18

Step-by-step explanation:

Given that:

P₁ = $10, P₂ = $20

From the tables Q₁ = 900, Q₂ = 800

Using midpoint method:

Percentage change in quantity = $\frac{Q_2-Q_1}{(Q_1+Q_2)/2} *100\%=\frac{800-900}{(900+800)/2}*100\%= -11.76\%\\$

Percentage change in price =

$\frac{P_2-P_1}{(P_1+P_2)/2} *100\%=\frac{20-10}{(20+10)/2}*100\%= 66.67\%\\$

Price of elastic demand = Percentage change in quantity/ Percentage change in price = -11.76% / 66.67% = 0.18

The Price of elastic demand is positive because we took the absolute value and elasticity are always positive

Therefore since Price of elastic demand < 1, the demand is inelastic in this interval.

This means that, along the demand curve between $10 to $20, if the price changes by 1%, the quantity demanded will change by 0.18%. A change in the price will result in a smaller percentage change in the quantity demanded. For example, a 10% increase in the price will result in only a 1.8% decrease in quantity demanded and a 10% decrease in the price will result in only a 1.8% increase in the quantity demanded

7 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

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

Hmmmmmmmmmmmmmmmmmmmmmmmmmmm

belka [17]

$\dfrac{6}{7}cd=\dfrac{6}{7}\cdot c\cdot d\\\\Factors:\ \dfrac{6}{7},\ c,\ d\\\\6x^2+7xy+3+9\\\\Constans\ (without\ x\ and\ y):\ 3,\ 9\\\\2x-4y+8=(2x)+(-4y)+(8)\\\\Terms:\ 2x,\ -4y,\ 8$