Explanation:
Find the two closest square numbers to 1659.
40*40=1600
41*41=1681
Decide much closer to either 1600 or 1681 is 1659.
If it is exactly halfway between the two numbers, the square root is 40.5.
If it is closer to 1600, the sq rt decimal is between 1 and 4.
If it is closer to 1681, the sq rt decimal is between 6 and 9.
we can simply pick any two points off the x,y table, so
![\bf \begin{array}{|cc|ll} \cline{1-2} x&y\\ \cline{1-2} 0&20\\ 3&11\\ 6&2\\ 9&-7\\ \cline{1-2} \end{array} \begin{array}{llll} \\[1em] \leftarrow \textit{let's use this point}\\\\ \leftarrow \textit{and this point} \end{array}~\hspace{5em} (\stackrel{x_1}{3}~,~\stackrel{y_1}{11})\qquad (\stackrel{x_2}{9}~,~\stackrel{y_2}{-7}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-7-11}{9-3}\implies \cfrac{-18}{6}\implies -3](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7B%7Ccc%7Cll%7D%0A%5Ccline%7B1-2%7D%0Ax%26y%5C%5C%0A%5Ccline%7B1-2%7D%0A0%2620%5C%5C%0A3%2611%5C%5C%0A6%262%5C%5C%0A9%26-7%5C%5C%0A%5Ccline%7B1-2%7D%0A%5Cend%7Barray%7D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5C%5C%5B1em%5D%0A%5Cleftarrow%20%5Ctextit%7Blet%27s%20use%20this%20point%7D%5C%5C%5C%5C%0A%5Cleftarrow%20%5Ctextit%7Band%20this%20point%7D%0A%5Cend%7Barray%7D~%5Chspace%7B5em%7D%0A%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B11%7D%29%5Cqquad%0A%28%5Cstackrel%7Bx_2%7D%7B9%7D~%2C~%5Cstackrel%7By_2%7D%7B-7%7D%29%0A%5C%5C%5C%5C%5C%5C%0Aslope%20%3D%20m%5Cimplies%0A%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B-7-11%7D%7B9-3%7D%5Cimplies%20%5Ccfrac%7B-18%7D%7B6%7D%5Cimplies%20-3)
Answer:
- 0.01688496
- 0.3449537
- 0.05002308
- 2.0375623
- 0.4162862
Step-by-step explanation:
<u>Some formulas to help</u>
- <em>log ab = log a + log b</em>
- <em>log a/b = log a - log b</em>
- <em>log a^b = b log a</em>
- <em>antilog (log a) = a, antilog is the inverse of log</em>
- <em>get values of log and antilog by using calculator or online calculator ( I used online calculator for this problem)</em>
- <em>round numbers as required, I left them as is</em>
1. Let the number be x, solving to show the method
- √0.0002851 = x
- log √0.0002851 = log x
- 1/2 log 0.0002851 = log x
- 1/2(-3.545) = log x
- log x = -1.7725
- antilog (log x) = antilog (-1.7725)
- x = 0.01688496
2. Short of the above method, will apply to this and following
=- antilog (1/7 log (0.0005812)) =
- antilog (1/7(-3.23567439437)) =
- antilog (-0.46223919919) =
- 0.3449537
3. .........................................
- 2.714^3 =
- antilog (log 2.714^3) =
- antilog (3 log 2.714) =
- antilog (3*0.43360984332) =
- antilog (1.30082952996) =
- 0.05002308
4..........................................
- 35.12^(1/5) =
- antilog (1/5 log (35.12)) =
- antilog (1/5*1.54555450723)
- antilog (0.30911090144) =
- 2.0375623
5. .........................................
- (0.07214)^(1/3) =
- antilog ( 1/3 log (0.07214)) =
- antilog (1/3*(-1.14182386202
)) =
- antilog ( --0.380607954
)=
- 0.4162862
Let me know if anything is not clear. Hope it helps.