To find a positive and a negative angle coterminal<span> with a given </span>angle<span>, you can add and subtract if the </span>angle<span> is measured in degrees or if the </span>angle<span> is measured in radians</span>
Answer:the error is when 6y=12
Step-by-step explanation:
the original problem says 4x-6y=12
so when you get rid of the 4x, it should still be -6y not positive
the yearly increase of x% assumes is compounding yearly, so let's use that.
![95000=80000\left(1+\frac{~~ \frac{r}{100}~~}{1}\right)^{1\cdot 5}\implies \cfrac{95000}{80000}=\left( 1+\cfrac{r}{100} \right)^5 \\\\\\ \cfrac{19}{16}=\left( 1+\cfrac{r}{100} \right)^5\implies \sqrt[5]{\cfrac{19}{16}}=1+\cfrac{r}{100}\implies \sqrt[5]{\cfrac{19}{16}}=\cfrac{100+r}{100} \\\\\\ 100\sqrt[5]{\cfrac{19}{16}}=100+r\implies 100\sqrt[5]{\cfrac{19}{16}}-100=r\implies 3.5\approx r](https://tex.z-dn.net/?f=95000%3D80000%5Cleft%281%2B%5Cfrac%7B~~%20%5Cfrac%7Br%7D%7B100%7D~~%7D%7B1%7D%5Cright%29%5E%7B1%5Ccdot%205%7D%5Cimplies%20%5Ccfrac%7B95000%7D%7B80000%7D%3D%5Cleft%28%201%2B%5Ccfrac%7Br%7D%7B100%7D%20%5Cright%29%5E5%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B19%7D%7B16%7D%3D%5Cleft%28%201%2B%5Ccfrac%7Br%7D%7B100%7D%20%5Cright%29%5E5%5Cimplies%20%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D1%2B%5Ccfrac%7Br%7D%7B100%7D%5Cimplies%20%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D%5Ccfrac%7B100%2Br%7D%7B100%7D%20%5C%5C%5C%5C%5C%5C%20100%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D%3D100%2Br%5Cimplies%20100%5Csqrt%5B5%5D%7B%5Ccfrac%7B19%7D%7B16%7D%7D-100%3Dr%5Cimplies%203.5%5Capprox%20r)
Answer:
-13 + j*2
Step-by-step explanation:
The additive inverse of a complex number x = a +j*b
is a number y, such that
x + y = 0
This means that
y = -x = - a - j*b
Therefore
The additive inverse of 13 - j*2 is equal to
-(13 - j*2) = -13 +j*2
Answer:
i think the second one or the first one
Step-by-step explanation:
hope this helps
