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4 years ago

NEED HELP FAST √−100= _ + _. WILL GIVE 90$

Mathematics

2 answers:

hoa [83]4 years ago

7 0

Answer:

0 + 10

Step-by-step explanation:

I'm for sure

Maslowich4 years ago

5 0

To get the answer for this question is:

√100/10 = √10

√10 + √90 (My $90) = √100

√100 + 444kesemealves = 0 + 10

Your answer is: 0 + 10.

Now give me my $90 ᕦ( ͡͡~͜ʖ ͡° )ᕤ

~ ShadowXReaper069

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3 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

4 years ago

At 107°F, a certain insect chirps at a rate of 92 times per minute, and at 113°F, they chirp 116 times per minute. Write an equa

dsp73

Answer:

The equation in slope-intercept form that represents the situation is y=0.25*x + 84 where y represents the temperature in ° F and x the number of chirps per minute.

Step-by-step explanation:

A linear equation can be expressed in the form y=m*x + b. In this equation, x and y are coordinates of a point, m is the slope and b is the y coordinate of the y-intercept. Since this equation describes a line in terms of its slope and its y-intercept, this equation is said to be in its slope-intercept form.

When there are two points of a line (x1, y1) and (x2, y2), the slope is determined by the quotient between the difference of the ordinate of these two points and the difference of the abscissa of the same points. This is:

$m=\frac{y2-y1}{x2-x1}$

Having a point on the line, you can substitute the values of m, x and y in the equation y = mx + b and thus find b.

In this case:

(x1, y1): (92, 107)
(x2, y2): (116, 113)

So:

$m=\frac{113-107}{116-92}$

m= 0.25

substituting the values of m, x1 and y1 in the equation y = mx + b you have:

107= 0.25*92 + b

107 - 0.25*92= b

84=b

The equation in slope-intercept form that represents the situation is y=0.25*x + 84 where y represents the temperature in ° F and x the number of chirps per minute.

3 0

3 years ago

Does anyone know the answer pleaseeee

pochemuha

Answer:

it b

Step-by-step explanation:

3 0

3 years ago

NEED HELP FAST √−100= ___ + ___. WILL GIVE 90$

NEED HELP FAST √−100= _ + _. WILL GIVE 90$