let's firstly convert the mixed fractions to improper fractions and then add them up, bearing in mind that the LCD from 12 and 6 will just be 12.
![\bf \stackrel{mixed}{14\frac{11}{12}}\implies \cfrac{14\cdot 12+11}{12}\implies \stackrel{improper}{\cfrac{179}{12}}~\hfill \stackrel{mixed}{3\frac{1}{6}}\implies \cfrac{3\cdot 6+1}{6}\stackrel{improper}{\cfrac{19}{6}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{179}{12}+\cfrac{19}{6}\implies \stackrel{\textit{using the LCD of 12}}{\cfrac{(1)179~~+~~(2)19}{12}}\implies \cfrac{179+38}{12}\implies \cfrac{217}{12}\implies 18\frac{1}{12}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B14%5Cfrac%7B11%7D%7B12%7D%7D%5Cimplies%20%5Ccfrac%7B14%5Ccdot%2012%2B11%7D%7B12%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B179%7D%7B12%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B3%5Cfrac%7B1%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B3%5Ccdot%206%2B1%7D%7B6%7D%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B19%7D%7B6%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B179%7D%7B12%7D%2B%5Ccfrac%7B19%7D%7B6%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20LCD%20of%2012%7D%7D%7B%5Ccfrac%7B%281%29179~~%2B~~%282%2919%7D%7B12%7D%7D%5Cimplies%20%5Ccfrac%7B179%2B38%7D%7B12%7D%5Cimplies%20%5Ccfrac%7B217%7D%7B12%7D%5Cimplies%2018%5Cfrac%7B1%7D%7B12%7D)
<span>3down votefavoriteFind the area between the circles <span><span><span>x2</span>+<span>y2</span>=4</span><span><span>x2</span>+<span>y2</span>=4</span></span> and <span><span><span>x2</span>+<span>y2</span>=6x</span><span><span>x2</span>+<span>y2</span>=6x</span></span> using polar coordinates.I have found that the equation of the first circle, call it <span><span>C1</span><span>C1</span></span>, is <span><span>r=2</span><span>r=2</span></span> on the other hand, for <span><span>C2</span><span>C2</span></span>, I get that its equation is <span><span>r=6cosθ</span><span>r=6cosθ</span></span>. Then, to find the bounds of integration, I have found that their angle of intersection should be <span><span>θ=arccos(1/3)</span><span>θ=arccos(1/3)</span></span> and <span><span>θ=−arccos(1/3)</span><span>θ=−arccos(1/3)</span></span>. Then, to set up the double integral:<span><span>A=<span><span>∫<span>arccos(1/3)</span><span>−arccos(1/3)</span></span><span><span>∫2<span>6cosθ</span></span>rdrdθ</span></span></span><span>A=<span><span>∫<span>−arccos(1/3)</span><span>arccos(1/3)</span></span><span><span>∫<span>6cosθ</span>2</span>rdrdθ</span></span></span></span>However, when evaluating this integral with the calculator, I get a negative value. What would be the problem in this case? Thanks in advance for your help.</span>
Calculator 1: Calculate the percentage of a number.
For example: 87% of 55 = 47.85
Calculator 2: Calculate a percentage based on 2 numbers.
For example: 47.85/55 = 87%
Answer:
2,205 cubic inches
Step-by-step explanation:
Answer:
they spent 20.3 hours every week watching television!
Step-by-step explanation:
Hope it helped :) please mark brainliest :)
