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3 years ago

which of the following are the building blocks of geometry, according to hilbert? ray line angle point plane

Mathematics

2 answers:

statuscvo [17]3 years ago

8 0

The building blocks of geometry would be point line and plane.

AleksAgata [21]3 years ago

8 0

Answer: line point plane

Step-by-step explanation:

We know that there are three undefined terms in geometry which we use to define other terms in geometry.

The three undefined terms in the geometry :-

a) Point b) Line c) Plane

According to Hilbert , the above three undefined terms in the geometry are the building blocks of geometry which are use to build the entire geometry .

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Maria and three friends have a thousand and two hundred stock cards if they share the stock of cards equally how many will each

andreev551 [17]

1200 ÷ 4 = 300
Each person will receive 300 stock cards.

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3 years ago

A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

3 years ago

When lava is expelled from a volcano, it temperature may vary from 700 degrees Celsius to 1200 degrees Celsius. Write an absolut

ValentinkaMS [17]

Answer:

Ix - 950°C I ≤ 250°C

Step-by-step explanation:

We are told that the temperature may vary from 700 degrees Celsius to 1200 degrees Celsius.

And that this temperature is x.

This means that the minimum value of x is 700°C while maximum of x is 1200 °C

Let's find the average of the two temperature limits given:

x_avg = (700 + 1200)/2 =

x_avg = 1900/2

x_avg = 950 °C

Now let's find the distance between the average and either maximum or minimum.

d_avg = (1200 - 700)/2

d_avg = 500/2

d_avg = 250°C.

Now absolute value equation will be in the form of;

Ix - x_avgI ≤ d_avg

Thus;

Ix - 950°C I ≤ 250°C

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2 years ago

The circumference of a circle can be found using the<br> formula C = 2 nr.

victus00 [196]

Radius = circumference/2(3.14)
OPTION 3

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Factor completely: x^2-y^2-1.5(x-y)

guapka [62]

Answer: x^2 - y ^2 − 1.5x + 1.5y

Step-by-step explanation:

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3 years ago

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