Since
The square root of 11 is somewhere between 3 and 4. In order to round it to the nearest tenth, we have to try all numbers between 3 and 4 with one decimal digit, and see which is closest to 11 when squared. We have
![\begin{array}{c|c}n&n^2\\3&9\\3.1&9.61\\3.2&10.24\\3.3&10.89\\3.4&11.56\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bc%7Cc%7Dn%26n%5E2%5C%5C3%269%5C%5C3.1%269.61%5C%5C3.2%2610.24%5C%5C3.3%2610.89%5C%5C3.4%2611.56%5Cend%7Barray%7D%5Cright%5D)
So, the square root of 11 is somewhere between 3.3 and 3.4.
If you add them all up it would be 6.
Answer:
64°
Step-by-step explanation:
∠RQP = 46°
∠PRQ = 180° - 110° = 70°
∠PRQ + ∠RQP + <u>∠RPQ</u> = 180°
70° + 46 + <u>∠RPQ</u> = 180°
116° + <u>∠RPQ</u> = 180°
Find <u>∠RPQ</u>
<u>∠RPQ</u> = 180° - 116° = 64°
Answer:
Range - {-4,0,12,20}
Step-by-step explanation:
Given that,
The function is :
g(x) = 4x –12
The domain of the function is {2, 3, 6, 8}.
g(2) = 4(2) –12 = -4
g(3) = 4(3) –12 = 0
g(6) = 4(6) –12 = 12
g(8) = 4(8) –12 = 20
Hence, the range of the function is {-4,0,12,20}.
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