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slamgirl [31]
3 years ago
9

What is another way of asking, "What is the log base of sixty four?"

Mathematics
1 answer:
shutvik [7]3 years ago
7 0
One number or word is missing in the statement.

Assume you mean "what is the log base 4 of 64?"

That is:

log_{4}64=x

Since, logarithm is the inverse function of exponential, another way to say that is to what exponent must be raised 4 to make 64?

This is, find x such that 4 ˣ = 64

Since 64 is 4³, the answer would be 3.

If the question were "what is the log base 2 of 64?" then other way to say it is to what number must be raised 2 to make 64?

And the answer is found making:

2ˣ = 64

2ˣ = 2⁶

Therefore, x = 6.
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On Call 3.5 hours: $10 Talk Time 1/2 hours: $1.25 What is the unit rate in dollars per hour for each company?
mafiozo [28]
On Call's unit rate is about $2.86 per hour, since $10/3.5 hours= $2.86/1 hour
Talk Time's unit rate is about $2.50 per hour, since $1.25/0.5 hours= $2.50/1 hour
3 0
3 years ago
I'd 11/20 closest to 1,1/2,or 0
timama [110]
It is closest to 1/2
7 0
3 years ago
Read 2 more answers
What is the markup of 23 to 41.40
makkiz [27]

Answer:

80%

Step-by-step explanation:

41.40/23 = 1.8 = 180% = 100% + 80%

The markup is 80%.

8 0
3 years ago
Find the area of the shaded region ​
o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)

A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill

\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}

5 0
2 years ago
by selling a mobile set of rupees 4800 Ramesh gains 25% how much will have we gained by selling it at rupees 4080​
Crank

Answer:

240

Step-by-step explanation:

Gain=25%

Gain=selling price - cost price

Gain = ((selling price - cost price )× 100)/ cost price

25c=(4800-c)100

25c=480000-100c

125c=480000

cost price = 3840

second statement

The selling price was 4080

Cost price 3840

therefore

Gain=selling price - cost price

Gain = 4080-3840

=240

5 0
3 years ago
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