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1 year ago

If Jose earns $1700 this year selling T-shirt’s he screen prints in his garage what taxes would he need to pay?

Mathematics

2 answers:

zalisa [80]1 year ago

4 0

Answer:

Jose is required to pay income tax and state sales tax

Step-by-step explanation:

Jose is required to pay income tax and state sales tax on his earnings of $1700.

The income tax goes to the the federal government which is 10% of the amount earned for the tax bracket of $0 to $9,700 for singles and $0 to $19,400 for married couples filling jointly.

Tax = 10% of $1700

Tax = 0.10*1700

Tax = $170

The state sales tax goes to the states and depends upon the taxing system of the state where Jose is selling the T-shirts.

Some states in the USA have zero state sales tax; Alaska, Delaware, Montana, Oregon etc. On the other hand, these states have the highest sales tax rates in the USA; Tennessee, Louisiana, Arkansas, Washington etc

Marta_Voda [28]1 year ago

3 0

Answer:

State sales and income tax

Step-by-step explanation:

He pays income tax based on the $1700 he got when running his own business of selling T-shirt’s he screen prints in his garage and he also needs to pay for the State sales tax on each T-shirt sold to the state that his business is located in.

Hope it will find you well

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A large tank is partially filled with 100 gallons of fluid in which 20 pounds of salt is dissolved. Brine containing 1 2 pound o

Valentin [98]

Answer:

47.25 pounds

Step-by-step explanation:

$\dfrac{dA}{dt}=R_{in}-R_{out}$

<u>First, we determine the Rate In</u>

Rate In=(concentration of salt in inflow)(input rate of brine)

$=(0.5\frac{lbs}{gal})( 6\frac{gal}{min})\\R_{in}=3\frac{lbs}{min}$

Change In Volume of the tank, $\frac{dV}{dt}=6\frac{gal}{min}-4\frac{gal}{min}=2\frac{gal}{min}$

Therefore, after t minutes, the volume of fluid in the tank will be: 100+2t

Rate Out=(concentration of salt in outflow)(output rate of brine)

$R_{out}=(\frac{A(t)}{100+2t})( 4\frac{gal}{min})\\\\R_{out}=\frac{4A(t)}{100+2t}$

Therefore:

$\dfrac{dA}{dt}=3-\dfrac{4A(t)}{100+2t}\\\\\dfrac{dA}{dt}=3-\dfrac{4A(t)}{2(50+t)}\\\\\dfrac{dA}{dt}=3-\dfrac{2A(t)}{50+t}\\\\\dfrac{dA}{dt}+\dfrac{2A(t)}{50+t}=3$

This is a linear differential equation in standard form, therefore the integrating factor:

$e^{\int \frac{2}{50+t}dt}=e^{2\ln|50+t|}=e^{\ln(50+t)^2}=(50+t)^2$

Multiplying the DE by the integrating factor, we have:

$(50+t)^2\dfrac{dA}{dt}+(50+t)^2\dfrac{2A(t)}{50+t}=3(50+t)^2\\\{(50+t)^2A(t)\}'=3(50+t)^2\\$Taking the integral of both sides\\\int \{(50+t)^2A(t)\}'= \int 3(50+t)^2\\(50+t)^2A(t)=(50+t)^3+C $ (C a constant of integration)\\Therefore:\\A(t)=(50+t)+C(50+t)^{-2}$

Initially, 20 pounds of salt was dissolved in the tank, therefore: A(0)=20

$20=(50+0)+C(50+0)^{-2}\\20-50=C(50)^{-2}\\C=-\dfrac{30}{(50)^{-2}} =-30X50^2=-75000$

Therefore, the amount of salt in the tank at any time t is:

$A(t)=(50+t)-75000(50+t)^{-2}$

After 15 minutes, the amount of salt in the tank is:

$A(15)=(50+15)-75000(50+15)^{-2}\\=47.25$ pounds$

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ra1l [238]

Answer:

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Step-by-step explanation:

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