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3 years ago

30 PTS, Need assistance!

2 answers:

10 0

The types of problems found on ancient papyri, books, and tablets focuses primarily on problems that relate to daily life. The solution methods for many ancient cultures are generally verbal, with mathematical statements written out in words. Why did most ancient cultures primarily write out their mathematical texts in words?

Answer:

Most ancient societies had symbols to represent numbers, but they did not have symbols to represent operations or unknown quantities. Thus, the problems and solutions to the problems had to be written in word form.

Alja [10]3 years ago

8 0

Answer:

The following are the correct statements regarding mathematics in the Egyptian, Chinese, and Babylonian cultures:

Both Egyptian and Chinese number systems use base 10.

All three cultures had unique symbolism to represent their numbers . (Earlier symbols were used to depict things)

There are two primary sources for ancient Egyptian mathematics

(The primary sources are the Rhind Papyrus and the Moscow Papyrus)

Step-by-step explanation:

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Romashka-Z-Leto [24]

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3 years ago

What are the values for the coefficients and constant term of 0=2+3x^2-5x

Usimov [2.4K]

Answer:

x=2/3 or x=1

I had this before

Is the right answer

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3 years ago

Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )

Zina [86]

Answer:

$\purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

Step-by-step explanation:

$\cos( \alpha ) = \frac{ \sqrt{6} }{6} = \frac{1}{ \sqrt{6} } \\ \\ \therefore \: \sin( \alpha ) = \sqrt{1 - { \cos}^{2} ( \alpha ) } \\ \\ = \sqrt{1 - \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} } \\ \\ = \sqrt{1 - {\frac{1}{ {6} }}} \\ \\ = \sqrt{ {\frac{6 - 1}{ {6} }}} \\ \\ \red{\sin( \alpha ) = \sqrt{ { \frac{5}{ {6} }}} } \\ \\ \tan( \alpha ) = \frac{\sin( \alpha ) }{\cos( \alpha ) } = \sqrt{5} \\ \\ \sin( \beta ) = \frac{ \sqrt{2} }{4} \\ \\ \implies \: \cos( \beta ) = \sqrt{ \frac{7}{8} } \\ \\ \tan( \beta ) = \frac{\sin( \beta ) }{\cos( \beta ) } = \frac{1}{ \sqrt{7} } \\ \\ \tan( \alpha - \beta ) = \frac{ \tan \alpha - \tan \beta }{1 + \tan \alpha . \tan \beta} \\ \\ = \frac{ \sqrt{5} - \frac{1}{ \sqrt{7} } }{1 + \sqrt{5} . \frac{1}{ \sqrt{7} } } \\ \\ = \frac{ \sqrt{35} - 1 }{ \sqrt{7} + \sqrt{5} } \\ \\ \purple{ \bold{ \tan( \alpha - \beta ) = 1.00701798}}$

8 0

3 years ago

Who knows about ratios

Kaylis [27]

I can help you. Just post the question

5 0

3 years ago

Aaron participates in a walkathon for charity. He has a sponsor who has pledged a base donation of $2 for the first mile he walk

tino4ka555 [31]

Answer:

total amount donated when aaron walks n miles = $2 + $3(n-1)

Step-by-step explanation:

Aaron participates in a walkathon for charity. He has a sponsor who has pledged a base donation of $2 for the first mile he walks and then a certain dollar amount for each additional mile he walks. Based on this pledge, if Aaron walks 8 miles, the sponsor will donate a total of $23 to the charity.

Write a formula that can be used to determine the amount of money this sponsor donates when Aaron walks n miles.

total amount sponsor pays = base donations + (additional miles walked after first mile x amount paid)

when Aaron walks 8 miles

additional miles = 8 - 1 = 7

23 = 2 + 7x

23 - 2 = 7x

x = 3

additional amount paid is $3

total amount donated when aaron walks n miles = $2 + $3(n-1)

5 0

3 years ago