Answer: 75 degrees
Step-by-step explanation:
75+75+30=180
6u + 3u = 18
9u = 18
u = 18/9
u = 2
The answer to the question
Aryabhata, also called Aryabhata I or Aryabhata the Elder, (born 476, possibly Ashmaka or Kusumapura, India), astronomer and the earliest Indian mathematician whose work and history are available to modern scholars. He is also known as Aryabhata I or Aryabhata the Elder to distinguish him from a 10th-century Indian mathematician of the same name. He flourished in Kusumapura—near Patalipurta (Patna), then the capital of the Gupta dynasty—where he composed at least two works, Aryabhatiya (c. 499) and the now lost Aryabhatasiddhanta.
Aryabhatasiddhanta circulated mainly in the northwest of India and, through the Sāsānian dynasty (224–651) of Iran, had a profound influence on the development of Islamic astronomy. Its contents are preserved to some extent in the works of Varahamihira (flourished c. 550), Bhaskara I (flourished c. 629), Brahmagupta (598–c. 665), and others. It is one of the earliest astronomical works to assign the start of each day to midnight.
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Aryabhatiya was particularly popular in South India, where numerous mathematicians over the ensuing millennium wrote commentaries. The work was written in verse couplets and deals with mathematics and astronomy. Following an introduction that contains astronomical tables and Aryabhata’s system of phonemic number notation in which numbers are represented by a consonant-vowel monosyllable, the work is divided into three sections: Ganita (“Mathematics”), Kala-kriya (“Time Calculations”), and Gola (“Sphere”).
Since the exponent (-3) is negative, flip the expression to: 1/(5n^4)^3.
Notice how the negative exponent becomes positive as you flip it.
Now evaluate the powers in the denominator: 1/((5^3)(n^4)^3
I separated the constant (5) from the variable (n) to show you how the powers are evaluated.
1/(5x5x5)(nxnxnxn)(nxnxnxn)(nxnxnxn)
—-> the power four means that there are 4 multiples of n in the parentheses. The power 3 corresponds to how many groups.
1/(125)(n^12)
= 1/125n^12
