Answer:
62.4 square centimeters
Step-by-step explanation:
The picture of the question in the attached figure
we know that
The surface area of the triangular pyramid is equal to the area of the triangular base plus the area of its three lateral triangular faces
In this problem the triangles are equilateral, that means, the
surface area is equal to the area of four congruent equilateral triangles
so
![A=4[\frac{1}{2}(b)(h)]](https://tex.z-dn.net/?f=A%3D4%5B%5Cfrac%7B1%7D%7B2%7D%28b%29%28h%29%5D)
we have

substitute
![A=4[\frac{1}{2}(6)(5.2)]=62.4\ cm^2](https://tex.z-dn.net/?f=A%3D4%5B%5Cfrac%7B1%7D%7B2%7D%286%29%285.2%29%5D%3D62.4%5C%20cm%5E2)
Answer:
C) 3pi>9
Step-by-step explanation:
Answer:
Step-by-step explanation:
Direction: Opens Up
Vertex:
(
3
,
−
4
)
Focus:
(
3
,
−
15
4
)
Axis of Symmetry:
x
=
3
Directrix:
y
=
−
17
4
Select a few
x
values, and plug them into the equation to find the corresponding
y
values. The
x
values should be selected around the vertex.
Tap for more steps...
x
y
1
0
2
−
3
3
−
4
4
−
3
5
0
Graph the parabola using its properties and the selected points.
Direction: Opens Up
Vertex:
(
3
,
−
4
)
Focus:
(
3
,
−
15
4
)
Axis of Symmetry:
x
=
3
Directrix:
y
=
−
17
4
x
y
1
0
2
−
3
3
−
4
4
−
3
5
0
Answer:
3x-2
Step-by-step explanation:
JK =JL+LM
5x-8=JL +(2x -6)
JL = 5x-8 -2x+6
JL = 3x-2
Answer:
V = 34,13*π cubic units
Step-by-step explanation: See Annex
We find the common points of the two curves, solving the system of equations:
y² = 2*x x = 2*y ⇒ y = x/2
(x/2)² = 2*x
x²/4 = 2*x
x = 2*4 x = 8 and y = 8/2 y = 4
Then point P ( 8 ; 4 )
The other point Q is Q ( 0; 0)
From these two points, we get the integration limits for dy ( 0 , 4 )are the integration limits.
Now with the help of geogebra we have: In the annex segment ABCD is dy then
V = π *∫₀⁴ (R² - r² ) *dy = π *∫₀⁴ (2*y)² - (y²/2)² dy = π * ∫₀⁴ [(4y²) - y⁴/4 ] dy
V = π * [(4/3)y³ - (1/20)y⁵] |₀⁴
V = π * [ (4/3)*4³ - 0 - 1/20)*1024 + 0 )
V = π * [256/3 - 51,20]
V = 34,13*π cubic units