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olga55 [171]
3 years ago
5

Is sextillion a real number?

Mathematics
1 answer:
user100 [1]3 years ago
7 0

Hi there!


Yes, sextillion is a real number, and a BIG one!


Did you know that sextillion has TWENTY-ONE zeros? Yes, twenty-one!

It looks something like this:


⇒ 1,000,000,000,000,000,000,000


Really big, right?


⇒ Sextillion, for many of you who may be watching, isn't really considered a joke because of part of its root word. Its just a number. :)


Hope this helps!

Message me if you need anything else! I'd be happy to help! :D

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A trucker handed in a ticket at a toll booth showing that in 2 hours she had covered 159 on a toll road with a posted speed limi
Nimfa-mama [501]

Answer:

overspeeding,

see if we calculate the speed or average speed,

it would be 159/2 which would be 79.5mph,

so as posted speed was 65mph, but the average speed was 79.5mph(Meaning most of the time she would have been above 65 mph), So,

She would have been charged with overspeeding

HOPE IT HELPS

BYE!

4 0
2 years ago
Joanne cannot decide which of two washing machines to buy. The selling price of each is ​$480 The first is marked down by ​%50 T
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i can not be cas  thae are the same

Step-by-step explanation:

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2 years ago
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a computer sales associate makes $74 each day that he works and makes approximately $20 in commission for each computer that he
podryga [215]

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your face

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3 years ago
What is being constructed based on the markings in the following diagram?
Liula [17]

Answer:

  perpendicular line through a point on a line

Step-by-step explanation:

The circle centered at C seems intended to produce point D at the same distance as point B. That is, C is the midpoint of BD.

The circles centered at B and D with radius greater than BC seems intended to produce intersection points G and H. (It appears accidental that those points are also on circle C. As a rule, that would be difficult to do in one pass.)

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Please help logarithms!
nlexa [21]

Given:

\log_34\approx 1.262

\log_37\approx 1.771

To find:

The value of \log_3\left(\dfrac{4}{49}\right).

Solution:

We have,

\log_34\approx 1.262

\log_37\approx 1.771

Using properties of log, we get

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_349      \left[\because \log_a\dfrac{m}{n}=\log_am-\log_an\right]

\log_3\left(\dfrac{4}{49}\right)=\log_34-\log_37^2      

\log_3\left(\dfrac{4}{49}\right)=\log_34-2\log_37          [\log x^n=n\log x]

Substitute \log_34\approx 1.262 and \log_37\approx 1.771.

\log_3\left(\dfrac{4}{49}\right)=1.262-2(1.771)

\log_3\left(\dfrac{4}{49}\right)=1.262-3.542

\log_3\left(\dfrac{4}{49}\right)=-2.28

Therefore, the value of \log_3\left(\dfrac{4}{49}\right) is -2.28.

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3 years ago
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