All categories

4 years ago

In a axiomatic system which category do pionts and lines and plane belong to?

Mathematics

2 answers:

hoa [83]4 years ago

8 0

Answer:

Undefined terms basicly was he said

Step-by-step explanation:

Grace [21]4 years ago

6 0

In an axiomatic system, points, lines, and planes belong to the category of undefined terms

You might be interested in

The solution of -2x + 13 <9 is

Karo-lina-s [1.5K]

Inequality form: x > 2
Interval notation: (2, ∞ )

8 0

3 years ago

A pen drops from the top of a desk that is 4 feet tall. The distance from the pen to the ground can be modeled by the equation b

Tcecarenko [31]

Answer:

0.5 seconds

Step-by-step explanation:

On the ground, height is 0

-16t² + 4 = 0

16t² = 4

t² = 4/16

t² = ¼

t = ½

4 0

3 years ago

3) The sum of three consecutive numbers is 72. What are the smallest of these numbers?

34kurt

Answer:

Step-by-step explanation:

23 + 24 + 25 = 72

6 0

3 years ago

Simplify. -(6x + 3y) - (x + y)

Arlecino [84]

Answer:

(-6x -3y ) -x-y

-6x-x-3y-y

5x +2y

7 0

3 years ago

We take a milk carton from the refrigerator and put it on the table at room temperature. We assume that the temperature of the c

Kisachek [45]

We can solve this problem using separation of variables.

Then apply the initial conditions

EXPLANATION

We were given the first order differential equation

$\frac{dT}{dt}=k(T-a)$

We now separate the time and the temperature variables as follows,

$\frac{dT}{T-a}=kdt$

Integrating both sides of the differential equation, we obtain;

$ln(T-a)=kt +c$

This natural logarithmic equation can be rewritten as;

$T-a=e^{kt +c}$

Applying the laws of exponents, we obtain,

$T-a=e^{kt}\times e^{c}$

$T-a=e^{c}e^{kt}$

We were given the initial conditions,

$T(0)=4$

Let us apply this condition to obtain;

$4-20=e^{c}e^{k(0)}$

$-16=e^{c}$

Now our equation, becomes

$T-a=-16e^{kt}$

or

$T=a-16e^{kt}$

When we substitute a=20,
we obtain,

$T=20-16e^{kt}$

b) We were also given that,

$T(5)=8$

Let us apply this condition again to find k.

$8=20-16e^{5k}$

This implied

$-12=-16e^{5k}$

$\frac{-12}{-16}=e^{5k}$

$\frac{3}{4}=e^{5k}$

We take logarithm to base e of both sides,

$ln(\frac{3}{4})=5k$

This implies that,

$\frac{ln(\frac{3}{4})}{5}=k$

$k=-0.2877$

After 15 minutes, the temperature will be,

$T=20-16e^{-0.2877\times 15}$

$T=20-0.21376$

$T=19.786$

After 15 minutes, the temperature is approximately 20°C

6 0

3 years ago