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2 years ago

Need help please snazzy fast

Mathematics

1 answer:

Alex777 [14]2 years ago

8 0

Answer:

-x and 6y

Step-by-step explanation:

They are present in the question.

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32.45 x 827.33 help??

natita [175]

26,846.8585 is the awnser

7 0

3 years ago

Read 2 more answers

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,

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

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 :

⇒ $d(t)=8sin(\pi t+c)$ ------1

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

⇒ $0=8sinc\\c=n\pi$ ------2

where n is an integer ;

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

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

where k is also an integer ;

the least value of $k=0$ ;

thus ,

the equation is :

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

7 0

3 years ago

5.6 x 5.6 x 6.16 x 6.16

ivann1987 [24]

Answer:

5.6 x 5.6 x 6.16 x 6.16=1190

Step-by-step explanation:

5.6 x 5.6 x 6.16 x 6.16=1189.974016

Rounded up will be 1190.

8 0

3 years ago

X+y=4 and x-y=6 substitution

vlabodo [156]

Y = -1
x = 5
jeng jeng jeng

4 0

3 years ago

Read 2 more answers

The circumference of a sphere was measured to be 80 cm with a possible error of 0.5 cm. A) Use differentials to estimate the max

siniylev [52]

Answer:

A) The maximum error in the calculated surface area: $25cm^2$

Relative error: $0.013$

B) The maximum error in the calculated volume: $162cm^2$

Relative error: $0.019$

Step-by-step explanation:

A) The formula for the surface area is:

$A=4\pi r^2$

The measured value is the circumference which is equal to:

$C=2\pi r$

then the radius is:

$r=\frac{C}{2\pi}$

Substituting in the formula of the surface:

$A=4\pi(\frac{C}{2\pi})^2\\A=4\pi(\frac{C^2}{4\pi^2})\\A=\frac{C^2}{\pi}$

Using the formula to calculate the error:

$dy=f'(x)dx$

Where $x$ is the variable measured and $y$ is a function of $x$ ( $y=f(x)$ ).

$dA=f'(C)dC\\dA=\frac{2C^{(2-1)}}{\pi}dC\\dA=\frac{2C}{\pi}dC$

We have C=80cm and dC=0.5cm

$dA=\frac{2C}{\pi}dC\\dA=\frac{2(80)}{\pi}(0.5)\\dA=\frac{160}{\pi}(0.5)\\dA=50.9296(0.5)\\dA=25.4648\approx25cm^2$

The relative error is the maximum error divide by the total area. The total area is: $A=\frac{C^2}{\pi}=\frac{(80)^2}{\pi}=\frac{6400}{\pi}=2037.1833cm^2$

$\frac{dA}{A}=\frac{25.4648}{2037.1833} =0.0125\approx0.013$

B) The formula for the volume is:

$V=\frac{4}{3} \pi r^3$

Using $r=\frac{C}{2\pi}$

$V=\frac{4}{3} \pi r^3\\V=\frac{4}{3} \pi (\frac{C}{2\pi})^3\\V=\frac{4}{3} \pi (\frac{C^3}{8\pi^3})\\V=\frac{1}{3}(\frac{C^3}{2\pi^2})\\V=\frac{C^3}{6\pi^2}$

The maximum error is:

$dV=\frac{3C^{3-1}}{6\pi^2}dC\\dV=\frac{C^{2}}{2\pi^2}dC\\dV=\frac{(80)^{2}}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=\frac{6400}{2\pi^2}(0.5)\\dV=(324.2278)(0.5)\\dV=162.1139\approx162cm^2$

The calculated volume is:

$V=\frac{C^3}{6\pi^2}\\V=\frac{(80)^3}{6\pi^2}\\V=\frac{512000}{6\pi^2}\\V=8646.0743$

The relative error is:

$\frac{dV}{V}=\frac{162.1139}{8646.0743}=0.0188\approx0.019$

3 0

3 years ago