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3 years ago

Can someone please help

Mathematics

1 answer:

Vsevolod [243]3 years ago

5 0

It may be convenient here to write the equation of the line as

... ∆y(x -x0) -∆x(y -y0) = 0

where (x0, y0) is a point on the line, and (∆x, ∆y) is the difference between two points on the line.

Using the first two points, we have

... (∆x, ∆y) = (-9, 10) - (-15, 5) = (6, 5)

So, the equation of the line is

... 5(x +9) -6(y -10) = 0 . . . . . . using the point (x0, y0) = (-9, 10)

... 5x -6y = -105 . . . . . . . . . . . simplify to standard form

Now, dividing by the constant on the right, we can put this into intercept form

... x/(-21) +y/(17.5) = 1

This tells us the y-intercept is (0, 17.5) and the x-intercept is (-21, 0).

_____

You can work directly with the values in the table. The y-value increases by 5 from one line to another, while the x-value increases by 6. Thus, to make the y-value decrease by 5 from the first point, we need to decrease that point's x-value by 6 to -21. That is, the x-intercept is (-21, 0).

Similarly, the x-value need to increase by only 3 from the last point. That amounts to half of the usual increase of 6. Thus the y-intercept will be the last point plus have the usual y-increase of 5, or 15+2.5 = 17.5. That is, the y-intercept is (0, 17.5).

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2 years ago

Two chemicals A and B are combined to form a chemical C. The rate, or velocity, of the reaction is proportional to the product o

Orlov [11]

Answer:

25.35 grams of C is formed in 14 minutes

after a long time , the limiting amount of C = 60g ,

A = 0 gram

and B = 30 grams; will remain.

Step-by-step explanation:

From the information given;

Let consider x(t) to represent the number of grams of compound C present at time (t)

It is obvious that x(0) = 0 and x(5) = 10 g;

And for x gram of C;

$\dfrac{2}{3}x$ grams of A is used ;

also $\dfrac{1}{3} x$ grams of B is used

Similarly; The amounts of A and B remaining at time (t) are;

$40 - \dfrac{2}{3}x$ and $50 - \dfrac{1}{3}x$

Therefore ; rate of formation of compound C can be said to be illustrated as ;

$\dfrac{dx}{dt }\propto (40 - \dfrac{2}{3}x)(50-\dfrac{1}{3}x)$

= $k \dfrac{2}{3}( 60-x) \dfrac{1}{3}(150-x)$

where;

k = proportionality constant.

= $\dfrac{2}{9}k (60-x)(150-x)$

By applying the separation of variable;

$\dfrac{1}{(60-x)(150-x)}dx= \dfrac{2}{9}k dt \\ \\ \\$

Solving by applying partial fraction method; we have:

$\{ \dfrac{1}{90(60-x)} - \dfrac{1}{90(150-x)} \}dx = \dfrac{2}{9}kdt$

$\dfrac{1}{90}(\dfrac{1}{x-150}-\dfrac{1}{x-60})dx =\dfrac{2}{9}kdt$

Taking the integral of both sides ; we have:

$\dfrac{1}{90}\int\limits(\dfrac{1}{x-150}- \dfrac{1}{x-60})dx= \dfrac{2}{9}\int\limits kdx$

$\dfrac{1}{90}(In(x-150)-In(x-60)) = \dfrac{2}{9}kt+C$

$\dfrac{1}{90}(In(\dfrac{x-150}{x-60})) = \dfrac{2}{9}kt+C$

$In( \dfrac{x-150}{x-60})= 20 kt + C_1 \ \ \ \ \ where \ \ C_1 = 90 C$

$\dfrac{x-150}{x-60}= Pe ^{20 kt} \ \ \ \ \ where \ \ P= e^{C_1}$

Applying the initial condition x(0) =0 to determine the value of P

Replace x= 0 and t =0 in the above equation.

$\dfrac{0-150}{0-60}= Pe ^{0}$

$\dfrac{5}{2}=P$

Thus;

$\dfrac{x-150}{x-60}=Pe^{20kt} \\ \\ \\ \dfrac{x-150}{x-60}=\dfrac{5}{2}e^{20kt} \\ \\ \\ 2x -300 =5e^{20kt}(x-60)$

$2x - 300 = 5xe^{20kt} - 300 e^{20kt} \\ \\ 5xe^{20kt} -2x = 300 e^{20kt} -300 \\ \\ x(5e^{20kt} -2) = 300 e^{20kt} -300 \\ \\ x= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

Applying the initial condition for x(7) = 15 , to find the value of k

Replace t = 7 into $x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(7)= \dfrac{300 e^{20k(7)}-300}{5e^{20k(7)}-2}$

$15= \dfrac{300 e^{140k}-300}{5e^{140k}-2}$

$75e^{140k}-30 ={300 e^{140k}-300}$

$225e^{140k}=270$

$e^{140k}=\dfrac{270}{225}$

$e^{140k}=\dfrac{6}{5}$

$140 k = In (\dfrac{6}{5})$

$k = \dfrac{1}{140}In (\dfrac{6}{5})$

k = 0.0013

Thus;

$x(t)= \dfrac{300 e^{20kt}-300}{5e^{20kt}-2}$

$x(t)= \dfrac{300 e^{20(0.0013)t}-300}{5e^{20(0.0013)t}-2}$

$x(t)= \dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2}$

The amount of C formed in 14 minutes is ;

$x(14)= \dfrac{300 e^{(0.026)14}-300}{5e^{(0.026)14}-2}$

x(14) = 25.35 grams

Thus 25.35 grams of C is formed in 14 minutes

NOW; The limiting amount of C after a long time is:

$\lim_{t \to \infty} = \lim_{t \to \infty} (\dfrac{300 e^{(0.026)t}-300}{5e^{(0.026)t}-2})$

$\lim_{t \to \infty} (\dfrac{300- 300 e^{(0.026)t}}{2-5e^{(0.026)t}})$

As; $\lim_{t \to \infty} e^{-20kt} = 0$

⇒ $\dfrac{300}{5}$

= 60 grams

Therefore as t → $\infty$ ; x = 60

and the amount of A that remain = $40 - \dfrac{2}{3}x$

= $40 - \dfrac{2}{3}(60)$

= 40 -40

=0 grams

The amount of B that remains = $50 - \dfrac{1}{3}x$

= $50 - \dfrac{1}{3}(60)$

= 50 - 20

= 30 grams

Hence; after a long time ; the limiting amount of C = 60g , and 0 g of A , and 30 grams of B will remain.

I Hope That Helps You Alot!.

5 0

3 years ago

Show all work to write the equations of the lines, representing the following conditions, in the form y = mx + b, where m is the

Kobotan [32]

Answer:

Part A) $y=\frac{4}{3}x+\frac{14}{3}$

Part B) $y=-\frac{3}{4}x+\frac{1}{2}$

Step-by-step explanation:

Part A) Passes through (−2, 2) and parallel to 4x − 3y − 7 = 0

we have

$4x-3y-7=0$

Isolate he variable y

$3y=4x-7$

$y=\frac{4}{3}x-\frac{7}{3}$

The slope of the given line is

$m=\frac{4}{3}$

Remember that

If two lines are parallel then their slopes are the same

therefore

The slope of the line parallel to the given line is also

$m=\frac{4}{3}$

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=\frac{4}{3}$

$point\ (-2,2)$

substitute

$2=\frac{4}{3}(-2)+b$

solve for b

$b=2+\frac{8}{3}$

$b=\frac{14}{3}$

therefore

$y=\frac{4}{3}x+\frac{14}{3}$

Part B) Passes through (−2, 2) and perpendicular to 4x − 3y − 7 = 0

we have

$4x-3y-7=0$

Isolate he variable y

$3y=4x-7$

$y=\frac{4}{3}x-\frac{7}{3}$

The slope of the given line is

$m=\frac{4}{3}$

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal

therefore

The slope of the line perpendicular to the given line is

$m=-\frac{3}{4}$

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=-\frac{3}{4}$

$point\ (-2,2)$

substitute

$2=-\frac{3}{4}(-2)+b$

solve for b

$b=2-\frac{3}{2}$

$b=\frac{1}{2}$

therefore

$y=-\frac{3}{4}x+\frac{1}{2}$

3 0

4 years ago