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4 years ago

What is the slope of a line parallel to the line -3x+5y=-10

Mathematics

2 answers:

lapo4ka [179]4 years ago

4 0

This looks familiar, right? Like slope-intercept.

y=mx+b, where m is the slope.

Let's right it like that.

5y=3x-10 (add 3x to both sides, since we want to isolate the y.)

Now divide by 5, since - again - we want to isolate the y.

y= 3/5x-10/5 >>>> y= 3/5x-2

I think the slope is 3/5.

If the line happens to be parallel, the slope is the same. If perpendicular, it is opposite reciprocal. So if it was perpendicular, the slope would be -5/3.

max2010maxim [7]4 years ago

3 0

The slope can be found re-writting the function in the general format s y=mx+n where m is known as slope. For this case:

-3x+5y=-10, trasponding the x term

5y=3x-10, dividing by 5

y=3x/5-10/5, solving finally have the general format:

y=3/5x-2

Then the slope is the number that precede the x unknow.

Solution. The slope is 3/5.

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3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

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Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

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$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

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$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

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