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STatiana [176]
3 years ago
13

Which polynomial correctly combines the like terms and expresses the given polynomial in standard form? 8mn5 – 2m6 + 5m2n4 – m3n

3 + n6 – 4m6 + 9m2n4 – mn5 – 4m3n3
Mathematics
1 answer:
MaRussiya [10]3 years ago
8 0

Answer:

-6m^{6}+n^{6}+7mn^{5}+14m^{2}n^{4}-5m^{3}n^{3}

Step-by-step explanation:

given polynomial is:

8mn^{5}-2m^{6}+5m^{2}n^{4}-m^{3}n^{3}+n^{6}-4m^{6}+9m^{2}n^{4}- mn^{5}-4m^{3}n^{3}

write similar terms together,

-2m^{6}-4m^{6}+n^{6}+8mn^{5}-mn^{5}+5m^{2}n^{4}+9m^{2}n^{4}-m^{3}n^{3}-4m^{3}n^{3}

-6m^{6}+n^{6}+7mn^{5}+14m^{2}n^{4}-5m^{3}n^{3}

Hence, polynomial in standard form -6m^{6}+n^{6}+7mn^{5}+14m^{2}n^{4}-5m^{3}n^{3}

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Answer:

Part A) sin(A)=\frac{2\sqrt{42}}{23}

Part B) cos(A)=\frac{19}{23}

Part C) tan(A)=\frac{2\sqrt{42}}{19}

Step-by-step explanation:

Part A) we know that

In the right triangle ABC of the figure the sine of angle A is equal to divide the opposite side angle A by the hypotenuse

so

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substitute the values

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Part B) we know that

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so

cos(A)=\frac{AC}{AB}

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so

tan(A)=\frac{BC}{AC}

substitute the values

tan(A)=\frac{2\sqrt{42}}{19}

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