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bearhunter [10]
4 years ago
12

Perform the following area application.

Mathematics
2 answers:
Alik [6]4 years ago
7 0

Answer:

C


Step-by-step explanation:

<em>You will cover halfway up each wall. Since the total height of the wall is 8 feet, you will cover half of that, which is 4 feet. </em>

Length is 12 feet and full length will be covered.


So area covered in 1 wall is 4*12=48 sq. ft.

There are 4 walls, so total area to be covered is 4*48=192 sq. ft.


Each panel board is 4 feet by 8 feet, so 1 panel board's area is 4*8=32 sq. ft.


So number of panel boards required to cover 192 sq. ft. with 32 sq. ft. boards is \frac{192}{32}=6

So we need 6 panels. Answer choice C is right.

galina1969 [7]4 years ago
4 0
8 x 12 = 96 sq ft 
<span>4 x 8 = 32 sqft </span>

<span>96 / 32 = 3 panels </span>

<span>3 x 4 = 12 panels. I'm not sure which answer that would be. :-( If any, I would select C.) 6 panels. </span>
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Length: (x - 4) meters

Step-by-step explanation:

x² - 16

x² - 4²

(x - 4)(x + 4)

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X = (L + 2k)/(x - b). You have the correct answer selected
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Find the imaginary part of\[(\cos12^\circ+i\sin12^\circ+\cos48^\circ+i\sin48^\circ)^6.\]
iren [92.7K]

Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6

First, we can expand this power using the binomial theorem:

(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}

After that, we can apply De Moivre's theorem to expand each summand:(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)

The final step is to find the common factor of i in the last expansion. Now:

x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6

=\binom{6}{0}(\cos(12)+i\sin(12))^6(\cos(48)+ i\sin(48))^0+\binom{6}{1}(\cos(12)+i\sin(12))^5(\cos(48)+ i\sin(48))^1+\binom{6}{2}(\cos(12)+i\sin(12))^4(\cos(48)+ i\sin(48))^2+\binom{6}{3}(\cos(12)+i\sin(12))^3(\cos(48)+ i\sin(48))^3+\binom{6}{4}(\cos(12)+i\sin(12))^2(\cos(48)+ i\sin(48))^4+\binom{6}{5}(\cos(12)+i\sin(12))^1(\cos(48)+ i\sin(48))^5+\binom{6}{6}(\cos(12)+i\sin(12))^0(\cos(48)+ i\sin(48))^6

=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))

The last part is to multiply these factors and extract the imaginary part. This computation gives:

Re x^6=\cos 72+6cos 60\cos 48-6\sin 60\sin 48+15\cos 96\cos 48-15\sin 96\sin 48+20\cos 36\cos 144-20\sin 36\sin 144+15\cos 24\cos 192-15\sin 24\sin 192+6\cos 12\cos 240-6\sin 12\sin 240+\cos 288

Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288

(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)

A calculator simplifies the imaginary part Im(x⁶) to 0

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Hey guys I really need help on this one!
Marina CMI [18]
I’m pretty sure it’s the last one. :)
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