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AleksandrR [38]
2 years ago
13

What is the difference between axiom and postulate theorem

Mathematics
1 answer:
nydimaria [60]2 years ago
6 0

Answer:

Axioms or postulates are universal truths. They cannot be proved. Theorem are statements which can be proved.

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A CD usually sells for $17.00. If the CD is 30% off, and sales tax is 8%, what is the total price of the CD, including tax?
Mademuasel [1]

Answer:

First find what the sale price of the CD is. 30% off means the CD is selling for 70% of the original cost ( 100%-30% = 70%)


Multiply the original cost by 70%:

17.00 x 0.70 = $11.90 ( sale price)

Now you need to multiply the sale price by the tax percentage to find the amount of tax:

11.90 x 0.08 = 0.95 ( tax)


Now add the sale price and the tax for the total cost:

11.90 + 0.95 = $12.85


6 0
2 years ago
Uma fábrica produz 3.967 lápis feitos com madeira reflorestada por dia. Se essa produção diária for mantida, quantos lápis serão
shtirl [24]

Responder:

111076 lápis

Explicação passo a passo:

Dado :

Número de lápis produzidos por dia = 3967 lápis

Se a produção diária for mantida, o número de lápis produzidos em 28 / dias será;

Produção diária * Número de dias

3967 * 28

= 111076 lápis

4 0
2 years ago
Where is y=-4/5x-1 located on the coordinate plane?
kramer
-4 left

+5 up - 1

Your welcome!
7 0
3 years ago
Read 2 more answers
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
Need some help on one question for geometry!
rjkz [21]
It’s 40 there you go
3 0
3 years ago
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