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Igoryamba
4 years ago
9

Help? A. 52°B. 180°C. 38°D. 128°

Mathematics
2 answers:
yawa3891 [41]4 years ago
8 0

<ACD and <DCB are supplementary so sum = 180

<ACD = 180 - <DCB

<ACD = 180 - 52

<ACD = 128

Answer

D. 128°

Arte-miy333 [17]4 years ago
4 0

The correct answer would be C.

Thank you hope this helps

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(a^(-5)b^(7)c^(3))^(2)-:(a^(4)b^(-2)c^(2))^(3) What are the powers of a b and c
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Answer:

The powers of a,b and c for the given expression

\frac{(a^{-5}b^7c^3)^2}{(a^4b^{-2}c^2)^3}=a^{-22}b^{20}c^0} is -22,20 and 0 respectively

Step-by-step explanation:

Given expression is  \frac{(a^{-5}b^7c^3)^2}{(a^4b^{-2}c^2)^3}

To find the powers of a,b and c:

\frac{(a^{-5}b^7c^3)^2}{(a^4b^{-2}c^2)^3}=\frac{(a^{-5})^2(b^7)^2(c^3)^2}{(a^4)^3(b^{-2})^3(c^2)^3} ( using the property (ab)^m=a^m.b^m )

=\frac{a^{-10}b^{14}c^6}{a^{12}b^{-6}c^6} ( using the property (a^m)^n=a^{mn} )

=a^{-10}b^{14}c^6.a^{-12}b^{6}c^{-6} ( using the property \frac{1}{a^m}=a^{-m} )

=a^{-10-12}b^{14+6}c^{6-6} ( using the property a^m.a^n=a^{m+n} )

=a^{-22}b^{20}c^{0}

Therefore\frac{(a^{-5}b^7c^3)^2}{(a^4b^{-2}c^2)^3}=a^{-22}b^{20}c^{0}

The powers of a,b and c for the given expression

\frac{(a^{-5}b^7c^3)^2}{(a^4b^{-2}c^2)^3}=a^{-22}b^{20}c^0} is -22,20 and 0 respectively

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3 years ago
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