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3 years ago

Please please help I need this for my test tomorrow

Mathematics

1 answer:

Airida [17]3 years ago

5 0

Answer:

12. It's a metaphor. The sky is being compared to a polished mirror.

13. It's a metaphor. It is describing the moon as a golden petal.

14. It's a simile. It is comparing his cheek to a rose in the snow.

15. It's a metaphor. It is describing one's soul as a gorge filled with echos of song which is like singing.

Step-by-step explanation:

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Derive Euler's formula e^(ix) - cos(x) + i*sin(x) by using taylor series expansion

gavmur [86]

Answer:

the answer is probably 5

5 0

3 years ago

Could anyone help, please?

Dmitry_Shevchenko [17]

First, find the probability of each event:

1) the probability that the spinner will land on a 7.

Since the spinner is split 4 equal sections and there is only 1 sector with 7, we can say the probability of getting a 7 is 1/4 as there is only 1 of 7 out of the total of 4 sections.

<em>and</em>

2) the probability that the spinner will land on B.

Since the spinner is split into 3 equal sections, and there is only 1 sector for B, we can say the probability of getting B is 1/3.

To find the probability of 2 events, we need to multiply the two probabilities.

1/4*1/3 = 1/12

So the answer is 1/12.

4 0

4 years ago

What do you get when you multiply 3x(8x-11)

Masteriza [31]

2.18 that will be ur answer

3 0

4 years ago

Read 2 more answers

Need SMART student.Please helps<br>

Blababa [14]

Answer:12

Step-by-step explanation:

Find the picture I attached to this answer. I solved it clearly.

4 0

3 years ago

Calculate s f(x, y, z) ds for the given surface and function. g(r, θ) = (r cos θ, r sin θ, θ), 0 ≤ r ≤ 4, 0 ≤ θ ≤ 2π; f(x, y, z)

Triss [41]

$g(r,\theta)=(r\cos\theta,r\sin\theta,\theta)\implies\begin{cases}g_r=(\cos\theta,\sin\theta,0)\\g_\theta=(-r\sin\theta,r\cos\theta,1)\end{cases}$

The surface element is

$\mathrm dS=\|g_r\times g_\theta\|\,\mathrm dr\,\mathrm d\theta=\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

and the integral is

$\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\int_0^{2\pi}\int_0^4((r\cos\theta)^2+(r\sin\theta)^2)\sqrt{1+r^2}\,\mathrm dr\,\mathrm d\theta$

$=\displaystyle2\pi\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac\pi4(132\sqrt{17}-\sinh^{-1}4)$

###

To compute the last integral, you can integrate by parts:

$u=r\implies\mathrm du=\mathrm dr$

$\mathrm dv=r\sqrt{1+r^2}\,\mathrm dr\implies v=\dfrac13(1+r^2)^{3/2}$

$\displaystyle\int_0^4r^2\sqrt{1+r^2}\,\mathrm dr=\frac r3(1+r^2)^{3/2}\bigg|_0^4-\frac13\int_0^4(1+r^2)^{3/2}\,\mathrm dr$

For this integral, consider a substitution of

$r=\sinh s\implies\mathrm dr=\cosh s\,\mathrm ds$

$\displaystyle\int_0^4(1+r^2)^{3/2}\,\mathrm dr=\int_0^{\sinh^{-1}4}(1+\sinh^2s)^{3/2}\cosh s\,\mathrm ds$

$\displaystyle=\int_0^{\sinh^{-1}4}\cosh^4s\,\mathrm ds$

$=\displaystyle\frac18\int_0^{\sinh^{-1}4}(3+4\cosh2s+\cosh4s)\,\mathrm ds$

and the result above follows.

4 0

4 years ago