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Stella [2.4K]
3 years ago
11

Compare the table and equation. X y 2 10 3 15 4 20 Equation: y = 4. 5x Which representation has the greatest slope? The equation

has the greatest slope. The table has the greatest slope. The table and equation have the same slope. Their slopes cannot be determined.
Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
8 0

The slope of the line gives the measure of the steepness and the direction.As the slope of the given table is 5 and slope of the given equation is 4.5. Thus the slope of the given table is more than the slope of the given equation. Hence the option B is the correct option.

Given information-

The given equation in the problem is,

y=4.5x

<h3>The slope of the line</h3>

The standard equation of the line can be given as,

y=mx+c

Where m is the slope of the line.

Compare the given equation with the standard form we get the value of the slope of the given equation is 4.5.

<h3>The slope of the table</h3>

The slope of the line can be given as,

m=\dfrac{y_2-y_1}{x_2-x_1}

m=\dfrac{15-10}{3-2}

m=5

The slope of the value given in the table is 5.

As the slope of the given table is 5 and slope of the given equation is 4.5. Thus the slope of the given table is more than the slope of the given equation. Hence the option B is the correct option.

Learn more about the slope of the line here;

brainly.com/question/2514839

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A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo
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Answer:

a) y(t)=50000-49990e^{\frac{-2t}{25}}

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Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is \frac{dy}{dt}=R_{i}-R_{o}, where R_{i} is the rate of salt entering and R_{o} is the rate of salt going outside.

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\frac{dy}{dt}+\frac{2y}{25}=4000, and using the integrating factor e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}, therefore  (\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}, we get   \frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}, after integrating both sides y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C, therefore y(t)= 50000 +Ce^{\frac{-2t}{25}}, to find C we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions y(0)=10, so 10= 50000+Ce^{\frac{-0*2}{25}}

10=50000+C\\C=10-50000=-49990

Finally we can write an expression for the amount of salt in the tank at any time t, it is y(t)=50000-49990e^{\frac{-2t}{25}}

b) The tank will overflow due Rin>Rout, at a rate of 80 L/min-40L/min=40L/min, due we have 500 L to overflow \frac{500L}{40L/min} =\frac{25}{2} min=t, so we can evualuate the expression of a) y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7, is the salt concentration when the tank overflows

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