Answer:
58
Step-by-step explanation:
58/3 gives a remainder of 1
58/4 gives a remainder of 2
58/5 gives a remainder of 3
Answer:
x = -1, and y = -2, giving you the answer (-1, -2)
Step-by-step explanation:
In the second equation, we're told that y is equal to 2x. With that, we can replace "y" in the first one with "2x", giving us:
-4x - 2x = 6
-6x = 6
x = -1
Now that we know the value of x, we can simply plug it into the first equation to find y:
-4x - y = 6
-4(-1) - y = 6
4 - y = 6
-4 + y = -6
y = -6 + 4
y = -2
So x = -1, and y = -2
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)
Now that we’ve learned how to solve word problems involving the sum of consecutive integers, let’s narrow it down and this time, focus on word problems that only involve finding the sum of consecutive even integers.
But before we start delving into word problems, it’s important that we have a good understanding of what even integers, as well as consecutive even integers, are.
Even Integers
We know that even numbers are integers that can be divided exactly or evenly by 22. Thus, the general form of the even integer nn, is n = 2kn=2k, where kk is also an integer.
In other words, since even numbers are the multiples of 22, we can represent an even integer nn by 2k2k, where kk is also an integer. So if we have the even integers 1010 and 1616,
