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4 years ago

Determine the equation of the line with the points (–13,10) and (16, 15).

1 answer:

3 0

$\bf (\stackrel{x_1}{-13}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{16}~,~\stackrel{y_2}{15}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{15-10}{16-(-13)}\implies \cfrac{5}{16+13}\implies \cfrac{5}{29}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-10=\cfrac{5}{29}[x-(-13)] \\\\\\ y-10=\cfrac{5}{29}(x+13)\implies y-10=\cfrac{5}{29}x+\cfrac{65}{29}\implies y=\cfrac{5}{29}x+\cfrac{65}{29}+10 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=\cfrac{5}{29}+\cfrac{355}{29}~\hfill$

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Evaluate (y - 9) + (3 + x), when y = 15 and x = 5

snow_lady [41]

Hi ;-)

$y=15 \ and \ x=5\\\\(y-9)+(3+x)=(15-9)+(3+5)=6+8=\boxed{14}$

6 0

3 years ago

Read 2 more answers

A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is

tresset_1 [31]

Answer:

a) $x(t) = 13*e^(^-^\frac{t}{100}^)$

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

$\frac{dx}{dt} =$ ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

$conc = \frac{x(t)}{1000}\frac{kg}{L}$

- The amount of salt leaving the tank per unit time can be determined from:

salt flow-out = conc * V( flow-out )

salt flow-out = $\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\$

salt flow-out = $\frac{x(t)}{100}\frac{kg}{min}$

- The ODE becomes:

$\frac{dx}{dt} = 0 - \frac{x}{100}$

- Separate the variables and integrate both sides:

$\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)$

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

$13 = C*e^0 = C$

- The solution to the ODE becomes:

$x(t) = 13*e^(^-^\frac{t}{100}^)$

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

$x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg$

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg

7 0

3 years ago

What is the y-intercept of the line that is the perpendicular bisector of the segment joining (-3,-4) and (5,10)? express you an

Novosadov [1.4K]

My geometry program shows the equation of the bisector to be
.. 4x +7y = 25
When x = 0, the y-value is 25/7 = 3 4/7

The y-intercept is 3 4/7.

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4 years ago

Harrison ordered 6 cases of pickles for his restaurants. Each case contained p jars of pickles. Harrison delivered an equal numb

natita [175]

6 cases ordered
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Split 7 ways, each restaurant gets (1/7) of the total order = 6p/7 jars.

I sure hope 'p' is a multiple of 7, or else Harrison might
need to deal with the number of pickles in each jar.

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3 years ago

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attashe74 [19]

Answer:

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3 years ago