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4 years ago

Which expression is equivalent to 28p^9 q^-5/12p^-6q7 ?

Mathematics

2 answers:

antoniya [11.8K]4 years ago

6 0

Answer:

the answer is b

Step-by-step explanation:

i got it right on edge 2020

Sonja [21]4 years ago

4 0

We are given expression: $\frac{28p^9q^{-5}}{12p^{-6}q^7}$ .

In order to simplify given expression, let us break it into smaller parts.

$\frac{28}{12} = \frac{7}{3}$

$\frac{p^9}{p^{-6}} =p^{9-(-6)} = p^{9+6} = p^{15}.$

$\frac{q^{-5}}{q^7} =\frac{1}{q^{7-(-5)}} = \frac{1}{q^{7+5}} = \frac{1}{q^{12}}$

Combining all terms together, we get

$\frac{7p^{15}}{3q^{12}} .$

<h3>Therefore, final expression is $\frac{7p^{15}}{3q^{12}}.$ </h3>

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edward wants to have $50,000 in 10 years for college. what single deposit would he need to make snow into an account that pays 4

soldi70 [24.7K]

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Step-by-step explanation:

Here Compound amount =$50,000

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$Amount=Principal(1+rate/365)^{365\times\ time}\\\Rightarrow\ Amount=Principal(1+4.3/365)^{365\times\ 10}\\\Rightarrow\ 50000=P(1+.011)^{3650}\\\Rightarrow\ 50000=P(1.011)^{3650}\\\Rightarrow\ 50000=P(13.911)...........(approx)\\\Rightarrow\ P=\frac{50000}{13.911}=3594......(approx)$

Therefore Principal amount deposited by Edward is $3594.

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3 years ago

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Find the general solution of the following differential equation. Primes denote derivatives with respect to x. 9 x (x+5y )y prim

Shalnov [3]

Answer:

$\frac{x}{5y} -lnxy = K$

Step-by-step explanation:

Given the differential equation

$9x(x+5y )y' = 9y(x -5y )\\x(x+5y )\frac{dy}{dx} = y(x -5y)\\\frac{dy}{dx} = \frac{y(x -5y)}{x(x+5y )} .\\let\ y = vx\\\frac{dy}{dx} =v+ x\frac{dv}{dx} \\v+ x\frac{dv}{dx} = \frac{vx(x -5vx)}{x(x+5vx )} \\v+ x\frac{dv}{dx} = \frac{v(1 -5v)}{(1+5v )} \\x\frac{dv}{dx} = \frac{v(1 -5v)}{(1+5v )} - v\\x\frac{dv}{dx} = \frac{v(1 -5v)-v(1+5v)}{(1+5v )} \\x\frac{dv}{dx} = \frac{-10v^{2} }{1+5v} \\\frac{dx}{x} = \frac{ 1+5v}{-10v^{2}}dv\\$

$\int\limits\frac{dx}{x} = \frac{-1}{10} \int\limits v^{-2} dv - \frac{1}{2}\int\limits \frac{dv}{v} \\lnx + C = \frac{1}{10v}-\frac{1}{2}lnv\\ since\ y =vx\\lnx+C = \frac{1}{10(\frac{y}{x} )}-\frac{1}{2}ln(\frac{y}{x} )\\lnx+C = \frac{x}{10y}-\frac{1}{2}ln(\frac{y}{x} )$

$2lnx +2C = \frac{x}{5y} - ln \frac{y}{x} \\\frac{x}{5y} - ln \frac{y}{x} -2lnx = 2C\\\frac{x}{5y} -(ln \frac{y}{x} +2lnx) = 2C\\\frac{x}{5y} -(ln \frac{y}{x} +lnx^{2} ) = 2C\\\\\frac{x}{5y} -lnxy = 2C\\\frac{x}{5y} -lnxy = K\ where \ K = 2C$

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3 years ago