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padilas [110]
4 years ago
12

When a 3-digit number is rounded to the nearest ten and to the nearest hundred the answer is the same. What is one possible numb

er that fits this rule?
Mathematics
2 answers:
leva [86]4 years ago
8 0
899

Rounded to nearest ten = 900

Rounded to nearest hundred = 900
masya89 [10]4 years ago
5 0

Answer:

There are many numbers like 798 , 696, 898 etc.

Step-by-step explanation:

When a 3-digit number is rounded to the nearest ten and to the nearest hundred the answer is the same. We have to tell few numbers.

There are many numbers that fit here, like 798 , 696, 898. In both cases, it rounds off to same numbers.

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4 years ago
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:

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(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)

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Answer:

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