Part A:
If (7^2)^x = 1 then the value of x is equal 0
Because when x = 0 the
(7^2)^0 = 1 ; 7^0 = 1
Any number with 0 exponent is always equal 0, except 0.
Part B:
If (7^0)^x = 1, then the value of x is all real number
Because when 0 multiplies any number, it is always equal 0. So then 7^0 is always equal 1.
Example
if x = 1 then (7^0)^1 = 7^0 = 1
if x = -1 then (7^0)^(-1) = 7^0 = 1
if x = 1000 then (7^0)^(1000) = 7^0 = 1
How to get answer for number 1: | 4+2i |
How to get answer for number 2: | 5-i |
Number 3 how to get answer: | -3i |
![\left|a+bi\right|\:=\sqrt{\left(a+bi\right)\left(a-bi\right)}=\sqrt{a^2+b^2}\\\mathrm{With\:}a=0,\:b=-3\\=\sqrt{0^2+\left(-3\right)^2}\\Refine\\=\sqrt{9}\\\sqrt{9}\\\mathrm{Factor\:the\:number:\:}\:9=3^2\\=\sqrt{3^2}\\\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^n}=a\\\sqrt{3^2}=3\\= 3](https://tex.z-dn.net/?f=%5Cleft%7Ca%2Bbi%5Cright%7C%5C%3A%3D%5Csqrt%7B%5Cleft%28a%2Bbi%5Cright%29%5Cleft%28a-bi%5Cright%29%7D%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5C%5C%5Cmathrm%7BWith%5C%3A%7Da%3D0%2C%5C%3Ab%3D-3%5C%5C%3D%5Csqrt%7B0%5E2%2B%5Cleft%28-3%5Cright%29%5E2%7D%5C%5CRefine%5C%5C%3D%5Csqrt%7B9%7D%5C%5C%5Csqrt%7B9%7D%5C%5C%5Cmathrm%7BFactor%5C%3Athe%5C%3Anumber%3A%5C%3A%7D%5C%3A9%3D3%5E2%5C%5C%3D%5Csqrt%7B3%5E2%7D%5C%5C%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%5Cquad%20%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%5C%5C%5Csqrt%7B3%5E2%7D%3D3%5C%5C%3D%203)
Conjugate. i hoped that helped
1/2 * 1 * (1 + 1) = 1
1/2 * 2 * (2 + 1) = 3
1/2 * 3 * (3 + 1) = 6
1/2 * 4 * (4 + 1) = 10
Best answer me please!
