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Hunter-Best [27]
3 years ago
9

Use words to explain how to solve this equation: (x-10)(x-4)=0

Mathematics
2 answers:
salantis [7]3 years ago
8 0

Answer: x = 10 <em>OR</em> x = 4.

Step-by-step explanation: If x = 10 or 4 then the last step of the equation will always be multiplying by 0 which equates to 0 as well.

if x = 10

(10-10)(10-4) -> (0)(6) -> 0 x 6 = 0

if x = 4

(4-10)(4-4) -> (-6)(0) -> -6 x 0 = 0

Anit [1.1K]3 years ago
6 0

Answer:

x = 10 \: and \: 4

Step-by-step explanation:

Hope it is helpful....

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Answer:

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Step-by-step explanation:

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3 years ago
Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea
Aleks [24]

Recall that variation of parameters is used to solve second-order ODEs of the form

<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>

so the first thing you need to do is divide both sides of your equation by <em>t</em> :

<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>

<em />

You're looking for a solution of the form

y=y_1u_1+y_2u_2

where

u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

and <em>W</em> denotes the Wronskian determinant.

Compute the Wronskian:

W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}

Then

u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t

u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}

The general solution to the ODE is

y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}

which simplifies somewhat to

\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}

4 0
2 years ago
A 500-gallon tank initially contains 220 gallons of pure distilled water. Brine containing 5 pounds of salt per gallon flows int
Wittaler [7]

Answer: The amount of salt in the tank after 8 minutes is 36.52 pounds.

Step-by-step explanation:

Salt in the tank is modelled by the Principle of Mass Conservation, which states:

(Salt mass rate per unit time to the tank) - (Salt mass per unit time from the tank) = (Salt accumulation rate of the tank)

Flow is measured as the product of salt concentration and flow. A well stirred mixture means that salt concentrations within tank and in the output mass flow are the same. Inflow salt concentration remains constant. Hence:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = \frac{d(V_{tank}(t) \cdot c(t))}{dt}

By expanding the previous equation:

c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt} + \frac{dV_{tank}(t)}{dt} \cdot c(t)

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c_{0} \cdot f_{in} - c(t) \cdot f_{out} = V_{tank}(t) \cdot \frac{dc(t)}{dt}

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\frac{dc(t)}{dt} + \frac{f_{out}}{V_{tank}} \cdot c(t) = \frac{c_0}{V_{tank}} \cdot f_{in}

The solution of this equation is:

c(t) = \frac{c_{0}}{f_{out}} \cdot ({1-e^{-\frac{f_{out}}{V_{tank}}\cdot t }})

The salt concentration after 8 minutes is:

c(8) = 0.166 \frac{pounds}{gallon}

The instantaneous amount of salt in the tank is:

m_{salt} = (0.166 \frac{pounds}{gallon}) \cdot (220 gallons)\\m_{salt} = 36.52 pounds

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