What's the equivalent to log3 729=6

Mathematics

1 answer:

Kaylis [27]3 years ago

3 0

What's the equivalent to log3 729=6

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When making a long distance call from a certain pay phone, the first three minutes of a call cost $2.10. After that, each additi

Sauron [17]

Answer: the number of minutes of long distance call that one can make is lesser than or equal to 12 minutes.

Step-by-step explanation:

Let x represent the number of minutes of long distance call that one makes.

The first three minutes of a call cost $2.10. After that, each additional minute or portion of a minute of that call cost $0.45. This means that if x minutes of long distance call is made, the total cost would be

2.10 + 0.45(x - 3)

Therefore, the inequality to find the number of minutes one can call long distance for $6.15 is expressed as

2.10 + 0.45(x - 3) ≤ 6.15

2.10 + 0.45x - 1.35 ≤ 6.15

0.75 + 0.45x ≤ 6.15

0.45x ≤ 6.15 - 0.75

0.45x ≤ 5.4

x ≤ 5.4/0.45

x ≤ 12

4 0

4 years ago

Suppose a and b are both non zero real numbers. Find real numbers c and d such that 1/a+ib= c+id

Thepotemich [5.8K]

$\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

Explanation

$\frac{1}{a+bi}=c+di$

Step 1

multiplicate by the conjugate

$\begin{gathered} \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(bi)^2} \\ \frac{1}{a+bi}\cdot\frac{a-bi}{a+bi}=\frac{a-bi}{(a+bi)(a-bi)}=\frac{a-bi}{a^2-(-b^2)}=\frac{a-bi}{a^2+b^2} \end{gathered}$

notice that

$\begin{gathered} \frac{1}{a+bi}=\frac{a-bi}{a^2+b^2}=\frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i \\ \frac{a}{a^2+b^2}-\frac{b}{a^2+b^2}i=c+di \\ so \\ \end{gathered}$ $\begin{gathered} c=\frac{a}{a^2+b^2} \\ d=\frac{-b}{a^2+b^2} \end{gathered}$

I hope this helsp you

6 0

1 year ago

Solve by factoring x^2-8x-9 = 0

3241004551 [841]

Answer:

the answer is ( x-9) (x+1) = 0

Step-by-step explanation:

more to the solution is x= -1, 9

4 0

3 years ago

Read 2 more answers