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stepan [7]
3 years ago
14

3.) An attorney charges a fixed fee of $250 for an initial meeting and $150 per hour for all hours

Mathematics
1 answer:
Rashid [163]3 years ago
5 0

Answer:

A → C = 150h + 250 ( where C is charge )

B → C = 175h + 150

for 26 hours

A → C = (150 × 26 ) + 250 = 3900 + 250 = $4150

B → C = (175 × 26 ) + 150 = 4550 + 150 = $4700

Attorney A is cheaper for 26 hours, thus better deal

Equate the 2 equations to find hours they charge the same

175h + 150 = 150h + 250 ( subtract 150h from both sides )

25h + 150 = 250 ( subtract 150 from both sides )

25h = 100 ( divide both sides by 25 )

h = 4 ← number of hours when charges for both are equal

Attorney A becomes a better deal at 5 hours

Step-by-step explanation:

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mel-nik [20]

substitute the ordered pair (2, 3 )  into the 2 equations

(3 × 2 ) + (4 × 3 ) = 6 + 12 = 18 → True

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When (2, 3 ) is substituted into the first equation, the equation is true.

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3 years ago
Simplify. Evaluate the numerical bases. 3^2 b^2 c^-4 *3^-1 b^-5 c^7
photoshop1234 [79]
Hope I can be of assistance!

3^2b^2c^{-4}\cdot \:3^{-1}b^{-5}c^7 \ \textgreater \  \mathrm{Apply\:exponent\:rule}:\ \:a^b\cdot \:a^c=a^{b+c}
3^{-1}\cdot \:3^2=\:3^{2-1}=\:3^1=\:3 \ \textgreater \  3b^{-5}b^2c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  b^{-5}b^2=\:b^{2-5}=\:b^{-3}
3b^{-3}c^{-4}c^7

\mathrm{Apply\:exponent\:rule}: \:a^b\cdot \:a^c=a^{b+c} \ \textgreater \  c^{-4}c^7=\:c^{-4+7}=\:c^3
3b^{-3}c^3

\mathrm{Apply\:exponent\:rule}: \:a^{-b}=\frac{1}{a^b} \ \textgreater \  b^{-3}=\frac{1}{b^3} \ \textgreater \  3\cdot \frac{1}{b^3}c^3

\mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} \ \textgreater \  \frac{1\cdot \:3c^3}{b^3}

Finally
\mathrm{Apply\:rule}\:1\cdot \:a=a \ \textgreater \  \frac{3c^3}{b^3}

Hope this helps!
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Answer:

Step-by-step explanation:

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