The height would be 36 because you have to divide 144 by 4
Notice the grid, the points are (-5, -2) and (4, -1), thus
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\ \quad \\ \begin{array}{ccccccccc} &&x_1&&y_1&&x_2&&y_2\\ % (a,b) &&(~{{ 4}} &,&{{ -1}}~) % (c,d) &&(~{{ -5}} &,&{{ -2}}~) \end{array}~~ % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ d=\sqrt{[-5-4]^2+[-2-(-1)]^2}\implies d=\sqrt{(-5-4)^2+(-2+1)^2} \\\\\\ d=\sqrt{(-9)^2+(-1)^2}\implies d=\sqrt{81+1}\implies d=\sqrt{82}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccccccc%7D%0A%26%26x_1%26%26y_1%26%26x_2%26%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%26%28~%7B%7B%204%7D%7D%20%26%2C%26%7B%7B%20-1%7D%7D~%29%20%0A%25%20%20%28c%2Cd%29%0A%26%26%28~%7B%7B%20-5%7D%7D%20%26%2C%26%7B%7B%20-2%7D%7D~%29%0A%5Cend%7Barray%7D~~%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%5B-5-4%5D%5E2%2B%5B-2-%28-1%29%5D%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-5-4%29%5E2%2B%28-2%2B1%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-9%29%5E2%2B%28-1%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B81%2B1%7D%5Cimplies%20d%3D%5Csqrt%7B82%7D)
Answer:
an = -8 +3(n-1)
an = -11 +3n
f(n) = -11 +3n
Step-by-step explanation:
–8, –5, –2, 1, 4, ...
This is an arithmetic sequence.
We are increasing by 3 each time
-8 +3 = -5
-5+3 = -2
-2 +3 = 1
The common difference is 3
The formula for an arithmetic sequence is
an = a1 +d (n-1)
an = -8 +3(n-1)
an = -8 +3n -3
an = -11 +3n
The nth term is -11 +3n
f(n) = -11 +3n
Answer:
x = 1/14
Step-by-step explanation:
You can work it as is by subtacting ln(14), then taking antilogs:
ln(x) = -ln(14)
x = 14^-1
x = 1/14
___
Or you can rewrite to a single log and then take antilogs:
ln(14x) = 0
14x = 1
x = 1/14 . . . . . divide by the coeffient of x
Answer:
<em><u>Surface</u></em><em><u> </u></em><em><u>area</u></em><em><u>=</u></em><em><u>4</u></em><em><u>8</u></em><em><u> </u></em><em><u>in</u></em><em><u>^</u></em><em><u>2</u></em>
<em><u>See</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>image</u></em><em><u> </u></em><em><u>for</u></em><em><u> </u></em><em><u>solution</u></em>
