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3 years ago

How can you use geometry figures to solve real world problems?

Mathematics

2 answers:

Alexxandr [17]3 years ago

6 0

Archetecture and basic building

inna [77]3 years ago

5 0

Architecture- based on mathematics and physics- more precise geometry. Figures were used long long ago, even before Christ. One of the biggest inventions, and the most contributing, and world known best inventions was a circle. That was the first step for machines. Geometry figures and angles are not just used in buildings, but in art- like sculptures (for example, before Christ in the Ancient Greece).

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The height of a cone-shaped container is 18 centimeters and its radius is 7 centimeters. Meg fills the container completely with

sasho [114]

Answer:

Step-by-step explanation:

5 0

3 years ago

Solve the system by elimination. (Please explain how to do it)

Naddika [18.5K]

Apply to Row 2 : Row 2 + Row 1

2x + 2y + 3z = 0

y + 4z = -3

2x + 3y + 3z = 5

Apply to Row 3: Row 3 - Row 1

2x + 2y + 3z = 0

y + 4z = -3

y = 5

Apply to Row 3: Row 3 - Row 2

2x + 2y + 3z = 0

y + 4z = -3

-4z = 8

Simplify rows

2x + 2y + 3z = 0

y + 4z = -3

z = -2

Note that the matrix is in echelon form now. The next steps are for back substitution.

Apply to Row 2: Row 2 - 4 Row 3

2x + 2y + 3z = 0

y = 5

z = -2

Apply to Row 1: Row 1 - 3 Row 3

2x + 2y = 6

y = 5

z = -2

Apply to Row 1: Row 1 - 2 Row 2

2x = -4

y = 5

z = 2

Simplify the rows

x = -2

y = 5

z = -2

3 0

3 years ago

Find the radius of a circle with a area of 212 squared

natulia [17]

Answer:

Step-by-step explanation:

3 0

3 years ago

10) Given that two tangent lines are constructed from the shared point A outside a circle with the center O and points of tangen

lora16 [44]

Those two angles are congruent.

_____
The angles are corresponding parts of congruent triangles (CPCT). Triangles OBA and OCA are right triangles congruent by the HL postulate. The "H" is segment OA. The "L" is the radius, OB or OC. Of course the angles at B and C are 90° as with any radius and tangent line.

5 0

3 years ago

Atmospheric pressure decreases by about 11.8% for every 1000 meters you climb. The pressure at sea level is 1013 millibars (a un

erica [24]

At sea level, the pressure is 1013, that's when the altitude is at 0, sea level, let's see

$\bf \textit{Periodic Exponential Decay}\\\\
A=I(1 - r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &1013\\
I=\textit{initial amount}\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &0\\
p=period\to &1000
\end{cases}
\\\\\\
1013=I(1-0.118)^{\frac{0}{1000}}\implies 1013=I\cdot 1\implies 1013=I$

so, the inital amount is 1013, when t = 0,

$\bf \textit{Periodic Exponential Decay}\\\\
A=I(1- r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1013\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &t\\
p=period\to &1000
\end{cases}
\\\\\\
A=1013(1-0.118)^{\frac{t}{1000}}\implies A=1013(0.882)^{\frac{t}{1000}}$

now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.

7 0

3 years ago