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wolverine [178]

2 years ago

6

A cylinder has how many flat surfaces

2 answers:

Svetllana [295]2 years ago

8 0

Answer:

A cylinder regularly has 2 flat surfaces. Which are both at the top and at the bottom, the rest of it is round.

Step-by-step explanation:

I hope it helps! Have a great day!

<em>JellyBeanie~</em>

Blababa [14]2 years ago

3 0

Answer:

2

Step-by-step explanation:

A cylinder has 2 flat surfaces.

I hope it helps! Have a great day!

Anygays-

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One golfer's scores for the season are 88, 90, 86, 89, 96, and 85. Another

Allisa [31]

Mean of the first golfer = 89

Mean of the second golfer = 87

Range of the first golfer = 11

Range of the second golfer = 8

Explanation:

First golfer scores 88, 90, 86, 89, 96 and 85.

Sum of the scores of first golfer = 88 + 90 + 86 + 89 + 96 + 85 = 534

Number of observation of first golfer = 6

$\text {Mean} = \frac{\text {Sum of the observation}}{\text {Number of observation}}$

Mean of the first golfer = $\frac{534}{6}=89$

Mean of the first golfer = 89

Range of the first golfer = Highest score – Lowest score

= 96 – 85

Range of the first golfer = 11

Second golfer scores 91, 86, 88, 84, 90 and 83.

Sum of the scores of second golfer = 91 + 86 + 88 + 84 + 90 + 83 = 522

Number of observation of second golfer = 6

$\text {Mean} = \frac{\text {Sum of the observation}}{\text {Number of observation}}$

Mean of the second golfer = $\frac{522}{6}=87$

Mean of the second golfer = 87

Range of the second golfer = Highest score – Lowest score

= 91 – 83

Range of the second golfer = 8

Comparing the golfer's skills:

Mean of first golfer is greater than Mean of second golfer. (i.e. 89 > 87)

Range of first golfer is greater than Range of second golfer. (i.e. 11 > 8)

Thus, first golfer have more skills than second golfer.

6 0

4 years ago

Work out m and c for the line:<br> 2x + 3y +1=0 <br><br>m=<br>c =

Sergio [31]

Answer:

m = - $\frac{2}{3}$ , c = - $\frac{1}{3}$

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

2x + 3y + 1 = 0 ( subtract 2x + 1 from both sides )

3y = - 2x - 1 ( divide terms by 3 )

y = - $\frac{2}{3}$ x - $\frac{1}{3}$ ← in slope- intercept form

with m = - $\frac{2}{3}$ and c = - $\frac{1}{3}$

4 0

3 years ago

A buoy floating in the sea is bobbing in simple harmonic motion with period 2 seconds and amplitude 8 in. Its displacement d fro

Mrac [35]

Answer:

The equation of the displacement $d$ as a function of time $t$ is :

$d(t)=8sin(\pi t+\pi )$

Step-by-step explanation:

Generally , A simple harmonic wave is a sinusoidal function that is it can be expressed in simple $sin$ or $cos$ terms.

Thus,

$d(t) = Asin(wt+c)$

is the general form of displacement of a SHM.

where,

<em> $d(t)$ is the displacement with respect to the mean position at any time $t$ </em>
<em> $A$ is amplitude </em>
<em> $w$ is the natural frequency of oscillation ( $rads^{-1}$ )</em>
<em> $c$ is the phase angle which indicates the initial position of the object in SHM ( $rad$ )</em>

given,

Time period ( $T$ ) = $2s$
$A=8$
The natural frequency ( $w$ ) and time period ( $T$ ) is :

$w=\frac{2\pi} {T}$

∴

$w$ = $\frac{2\pi }{2} = \pi$ $rads^{-1}$

∴

the equation :

<em>⇒ $d(t)=8sin(\pi t+c)$ </em> ------1

since $d=0$ when $t=o$ ,

<em>⇒ $0=8sinc\\c=n\pi$ ------2</em>

where n is an integer ;

<u>⇒since the bouy immediately moves in the negative direction , $x$ must be negative or c must be an odd multiple of $\pi$ .</u>

<em>⇒ $d(t) = 8sin(\pi t+(2k+1)\pi )$ ------3</em>

where k is also an integer ;

the least value of $k=0$ ;

thus ,

the equation is :

$d(t)=8sin(\pi t+\pi )$

<em></em>

7 0

3 years ago

Solve the inequality for u<br>-5u+27 [less than or equal to] 2

baherus [9]

Answer:

5≤u

Step-by-step explanation:

Put the words and numbers into an equation:

-5u+27≤2

Put all like terms on one side:

27-2≤5u ---> Do you see what I did there? Now you don't have to work with negatives. :)

Simplify:

25≤5u

Solve:

5≤u

:)

6 0

3 years ago

kifflom [539]

$\\ \sf\longmapsto \dfrac{x^2-1}{x^2+5x+4}\leqslant 0$

Solving denominator

$\\ \sf\longmapsto x^2+5x+4>0$

$\\ \sf\longmapsto x^2+4x+x+4>0$

$\\ \sf\longmapsto x(x+4)+1(x+4)>0$

$\\ \sf\longmapsto (x+4)(x+1)>0$

$\\ \sf\longmapsto x>-4\:or x>-1$

Hence x $\neq$ -1
x can't be 0 as it makes function undefined

$\\ \sf\longmapsto -4$

5 0

3 years ago