1 white shirt
1 black shirt
1 blue shirt
1 blue pant
1 tan pant
There are 5 total articles of clothing and 2 that he could end up wearing (white shirt & tan pants)
so the probability equals 2/5
7x+y=11 where y= -5 So, when we substitute y in the equation then,
7x + (-5) = 11 that is 7x - 5 = 11....
So, B. is the right answer.
Rationalizing the denominator, simply means "getting rid of that pesky root at the bottom", and we do so by simply multiplying it by something to take it out, of course, we multiply the bottom, we have to also multiply the top,
![\bf \cfrac{4\sqrt{150}}{\sqrt{189x}}\cdot \cfrac{\sqrt{189x}}{\sqrt{189x}}\implies \cfrac{4\sqrt{150}\sqrt{189x}}{(\sqrt{189x})^2}\implies \cfrac{4\sqrt{(150)({189x})}}{189x}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B4%5Csqrt%7B150%7D%7D%7B%5Csqrt%7B189x%7D%7D%5Ccdot%20%5Ccfrac%7B%5Csqrt%7B189x%7D%7D%7B%5Csqrt%7B189x%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Csqrt%7B150%7D%5Csqrt%7B189x%7D%7D%7B%28%5Csqrt%7B189x%7D%29%5E2%7D%5Cimplies%20%5Ccfrac%7B4%5Csqrt%7B%28150%29%28%7B189x%7D%29%7D%7D%7B189x%7D)
1)
![(-2+\sqrt{-5})^2\implies (-2+\sqrt{-1\cdot 5})^2\implies (-2+\sqrt{-1}\sqrt{5})^2\implies (-2+i\sqrt{5})^2 \\\\\\ (-2+i\sqrt{5})(-2+i\sqrt{5})\implies +4-2i\sqrt{5}-2i\sqrt{5}+(i\sqrt{5})^2 \\\\\\ 4-4i\sqrt{5}+[i^2(\sqrt{5})^2]\implies 4-4i\sqrt{5}+[-1\cdot 5] \\\\\\ 4-4i\sqrt{5}-5\implies -1-4i\sqrt{5}](https://tex.z-dn.net/?f=%28-2%2B%5Csqrt%7B-5%7D%29%5E2%5Cimplies%20%28-2%2B%5Csqrt%7B-1%5Ccdot%205%7D%29%5E2%5Cimplies%20%28-2%2B%5Csqrt%7B-1%7D%5Csqrt%7B5%7D%29%5E2%5Cimplies%20%28-2%2Bi%5Csqrt%7B5%7D%29%5E2%20%5C%5C%5C%5C%5C%5C%20%28-2%2Bi%5Csqrt%7B5%7D%29%28-2%2Bi%5Csqrt%7B5%7D%29%5Cimplies%20%2B4-2i%5Csqrt%7B5%7D-2i%5Csqrt%7B5%7D%2B%28i%5Csqrt%7B5%7D%29%5E2%20%5C%5C%5C%5C%5C%5C%204-4i%5Csqrt%7B5%7D%2B%5Bi%5E2%28%5Csqrt%7B5%7D%29%5E2%5D%5Cimplies%204-4i%5Csqrt%7B5%7D%2B%5B-1%5Ccdot%205%5D%20%5C%5C%5C%5C%5C%5C%204-4i%5Csqrt%7B5%7D-5%5Cimplies%20-1-4i%5Csqrt%7B5%7D)
3)
let's recall that the conjugate of any pair a + b is simply the same pair with a different sign, namely a - b and the reverse is also true, let's also recall that i² = -1.
![\cfrac{6-7i}{1-2i}\implies \stackrel{\textit{multiplying both sides by the denominator's conjugate}}{\cfrac{6-7i}{1-2i}\cdot \cfrac{1+2i}{1+2i}\implies \cfrac{(6-7i)(1+2i)}{\underset{\textit{difference of squares}}{(1-2i)(1+2i)}}} \\\\\\ \cfrac{(6-7i)(1+2i)}{1^2-(2i)^2}\implies \cfrac{6-12i-7i-14i^2}{1-(2^2i^2)}\implies \cfrac{6-19i-14(-1)}{1-[4(-1)]} \\\\\\ \cfrac{6-19i+14}{1-(-4)}\implies \cfrac{20-19i}{1+4}\implies \cfrac{20-19i}{5}\implies \cfrac{20}{5}-\cfrac{19i}{5}\implies 4-\cfrac{19i}{5}](https://tex.z-dn.net/?f=%5Ccfrac%7B6-7i%7D%7B1-2i%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultiplying%20both%20sides%20by%20the%20denominator%27s%20conjugate%7D%7D%7B%5Ccfrac%7B6-7i%7D%7B1-2i%7D%5Ccdot%20%5Ccfrac%7B1%2B2i%7D%7B1%2B2i%7D%5Cimplies%20%5Ccfrac%7B%286-7i%29%281%2B2i%29%7D%7B%5Cunderset%7B%5Ctextit%7Bdifference%20of%20squares%7D%7D%7B%281-2i%29%281%2B2i%29%7D%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B%286-7i%29%281%2B2i%29%7D%7B1%5E2-%282i%29%5E2%7D%5Cimplies%20%5Ccfrac%7B6-12i-7i-14i%5E2%7D%7B1-%282%5E2i%5E2%29%7D%5Cimplies%20%5Ccfrac%7B6-19i-14%28-1%29%7D%7B1-%5B4%28-1%29%5D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B6-19i%2B14%7D%7B1-%28-4%29%7D%5Cimplies%20%5Ccfrac%7B20-19i%7D%7B1%2B4%7D%5Cimplies%20%5Ccfrac%7B20-19i%7D%7B5%7D%5Cimplies%20%5Ccfrac%7B20%7D%7B5%7D-%5Ccfrac%7B19i%7D%7B5%7D%5Cimplies%204-%5Ccfrac%7B19i%7D%7B5%7D)
Answer:
489.5 + m = 756.80 (the amount elias has, added to marks is equal to the total
(756.8- 489.5 = 267.30 which is the amount mark has.)
m= 267.3
the last option is the correct option
Step-by-step explanation: