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cestrela7 [59]
3 years ago
8

4

Mathematics
1 answer:
barxatty [35]3 years ago
4 0
The answer is 154cm^2
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A graph is shown below:
Alecsey [184]
10 and so what is the
4 0
3 years ago
There are 420 students in kindergarten through third grade if 1/4 of the students are in first grade 20% are in the second grade
Oduvanchick [21]
The answer would be around 120
5 0
3 years ago
Read 2 more answers
NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

8 0
2 years ago
Given f(x) = log(x+1), x &gt;-1 and g(x) = x^2 + 2x, XER find (f•g)(1)​
worty [1.4K]

Answer:

Like terms, functions may be combined by addition, subtraction, multiplication or division.

Example 1. Given f ( x ) = 2x + 1 and g ( x ) = x2

+ 2x – 1 find ( f + g ) ( x ) and

( f + g ) ( 2 )

Solution

Step 1. Find ( f + g ) ( x )

Since ( f + g ) ( x ) = f ( x ) + g ( x ) then;

( f + g ) ( x ) = ( 2x + 1 ) + (x2

+ 2x – 1 )

= 2x + 1 + x2

+ 2x – 1

= x

2

+ 4x

Step 2. Find ( f + g ) ( 2 )

To find the solution for ( f + g ) ( 2 ), evaluate the solution above for 2.

Since ( f + g ) ( x ) = x2

+ 4x then;

( f + g ) ( 2 ) = 22

+ 4(2)

= 4 + 8

= 12

Example 2. Given f ( x ) = 2x – 5 and g ( x ) = 1 – x find ( f – g ) ( x ) and ( f – g ) ( 2 ).

Solution

Step 1. Find ( f – g ) ( x ).

( f – g ) ( x ) = f ( x ) – g ( x )

= ( 2x – 5 ) – ( 1 – x )

= 2x – 5 – 1 + x

= 3x – 6

Step 2. Find ( f – g ) ( 2 ).

( f – g ) ( x ) = 3x – 6

( f – g ) ( 2 ) = 3 (2) – 6

= 6 – 6

= 0

Example 3. Given f ( x ) = x2

+ 1 and g ( x ) = x – 4 , find ( f g ) ( x ) and ( f g ) ( 3 ).

Solution

Step 1. Solve for ( f g ) ( x ).

Since ( f g ) ( x ) = f ( x ) * g ( x ) , then

= (x2

+ 1 ) ( x – 4 )

= x

3

– 4 x2

+ x – 4 .

Step 2. Find ( f g ) ( 3 ).

Since ( f g ) ( x ) = x3

– 4 x2

+ x – 4, then

( f g ) ( 3 ) = (3)3

– 4 (3)2

+ (3) – 4

= 27 – 36 + 3 – 4

= -10

Example 4. Given f ( x ) = x + 1 and g ( x ) = x – 1 , find ( x ) and ( 3 ). f

g

⎛ ⎞ ⎜

⎝ ⎠

f

g

⎛ ⎞ ⎜

⎝ ⎠ ⎟ ⎟

Solution

Step 1. Solve for ( x ). f

g

⎛

⎜

⎝ ⎠

⎞

⎟

Since ( x ) = , then ( )

( )

f x

g x

f

g

⎛

⎜

⎝ ⎠

⎞

⎟

= ; x ≠ 1 1

1

x

x

+

−

Step 2 Find . ( ) 3 f

g

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

Since = , then 1

1

x

x

+

− ( ) f x

g

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

=

3 1

3 1

+

− ( ) 3 f

g

⎛ ⎞ ⎜ ⎟ ⎝ ⎠

=

4

2

= 2

Step-by-step explanation:

did this Help?

3 0
3 years ago
Miles ordered 126 books to give away at the store opening. What is 126 rounded to the nearest hundred?
dlinn [17]
Miles ordered 126 = rounded : 130
3 0
3 years ago
Read 2 more answers
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