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)
John spent 30 on six cars 5 times 6 = 30
The answer is 74 i think so
Answer:
I would say that by the 20th figure, there would be more than 400 small triangles by the 20th figure.
Step-by-step explanation:
Based on figure 1 and 2, I'm guessing the ratio is 2.
Figure 4: 32
Figure 5: 64
Figure 6: 128
Figure 7: 256
Figure 8: 512
Figure 9: 1,024
Answer:
formula: a squared + b squared= c squared
6 squared+ 8 squared= c squared
36+64= c squared
100= c squared
Square root both sides
You get 10= c
