Answer: d
Step-by-step explanation: when you bring 3 into the square root, it becomes isqrt(63). when you apply i, it makes the number sqrt(-63)
pretty sure I did it right this time
The answer is 54 degrees I took the exam I got it right
Answer:
.
Step-by-step explanation:
Probability that the car encounters a green light on the first day:
.
To meet the question's conditions, the car needs to encounter another green light on the second day. Given that the colors of the light on each day are "independent," the chance that there's a green light followed by another green light will be
.
- Condition is met on the first two days and green light on the third day:
. - Condition is met on the first three days and green light on the fourth day:
.
To meet the condition on the fifth day, there needs to be a yellow light. The probability that the condition is met on the first four days and on the fifth day will be
.
To meet the condition on the sixth day, all prior days should meet the conditions. Besides, there needs to be a red light on the sixth day. 
- Seventh day:

- Eighth day:
- Ninth day:
- Tenth day:
The question asks that the condition be met on all ten days. As a result, the probability of meeting the condition will be equal to the probability on the tenth day:
.
Answer:
Circumcenter theorem states that the vertices of the each triangle are equidistant from the circumcenter.
As per the statement:
It is given that: P is the circumference
From the given figure:
CP = 12 units.
then;
by circumcenter theorem;
AP= BP =CP = 12 units.
Next find the value of AB:
Labelled the diagram:
AD = 11 units
then;
AB = AD+DB
Since: AD=DB [You can see it from the given figure]
then;
AB = 2AD = 2(11) = 22 units
Therefore, the value of BP and AB are: 12 units and 22 units
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)