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solong [7]
4 years ago
10

Someone please answer this. Please give answer in form of square root.

Mathematics
1 answer:
sergey [27]4 years ago
4 0
V=A_bh\\
A_b=\dfrac{3a^2\sqrt3}{2}\\
a=4\\
h=9\\\\
V=\dfrac{3\cdot4^2\sqrt3}{2}\cdot9\\
V=216\sqrt3
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A cable installer charges $40.00 per hour plus a $60.00 service charge. Your father's firm hires him to hook up his company's in
Vadim26 [7]
Set up the equation ... 60 + 40X = total charges... where X is the number of hours (8.5)

60 + 40(8.5) = ?
60 + 340 = ?
400 = total charges
4 0
3 years ago
Read 2 more answers
Plz no link i need real answers please
leonid [27]

Answer:

You can download the link here:

https://rb.gy/tmu0gl

JKJKJKJKJKJKJKJKJKJKJK

Your answer is n = 9/8

Brainliest? Thx.

3 0
3 years ago
7731 rounded to the nearest 10
Lana71 [14]
7731 Rounded to the nearest 10th is 7,731.0
7 0
3 years ago
Given the function f)(x) =-x^2+3x+8,determine the average rate of change of the function over the interval listed in the picture
Nadusha1986 [10]

Answer:

Vertex:

(

3

2

,

−

41

4

)

(

3

2

,

-

41

4

)

Focus:

(

3

2

,

−

10

)

(

3

2

,

-

10

)

Axis of Symmetry:

x

=

3

2

x

=

3

2

Directrix:

y

=

−

21

2

y

=

-

21

2

x

y

0

−

8

1

−

10

3

2

−

41

4

3

−

8

4

−

4

Vertex:

(

3

2

,

−

41

4

)

(

3

2

,

-

41

4

)

Focus:

(

3

2

,

−

10

)

(

3

2

,

-

10

)

Axis of Symmetry:

x

=

3

2

x

=

3

2

Directrix:

y

=

−

21

2

y

=

-

21

2

x

y

0

−

8

1

−

10

3

2

−

41

4

3

−

8

4

−

4

Step-by-step explanation:

7 0
3 years ago
A mass oscillating up and down on the bottom of a spring (assuming perfect elasticity and no friction or air resistance) can be
alexdok [17]

The motion of the mass as it moves on the bottom of the spring is a

repetitive motion.

  • \mathrm{The \ motion \ of \ the \ mass \ attached \ to \ the \ spring \ is \ }\displaystyle d = 6 \cdot sin\left(\frac{\pi}{2} \cdot t \right)

Reasons:

The general form of the equation of the simple harmonic motion of the

mass is <em>d</em> = a·sin(ω·t)

Where;

d = The distance of the mass from the rest position

a = The maximum displacement of the mass from the equilibrium position = 6 cm

ω = The frequency of rotation

t = The time of motion

ω = The frequency of rotation

\displaystyle \omega = \mathbf{\frac{2 \cdot \pi}{T}}

Where;

T = The time to complete one cycle (the period of oscillation) = 4 seconds

\displaystyle \omega = \frac{2 \cdot \pi}{4} = \frac{\pi}{2}

Combining the above values gives the modelling equation as follows;

\displaystyle d = 6 \cdot sin\left(\frac{\pi}{2} \cdot t \right)

Learn more here:

brainly.com/question/12904891

3 0
2 years ago
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