Answer:
A point on either side of the line
Step-by-step explanation:
To determine which side to shade, you look at a point on each side of the line. One will make the inequality true and one will not. Shade on the side that will make the inequality true
Answer:
option c boo if it's for plata
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)
Choice D is true. It is just another way to write the given function.
To confirm choices A through C, replace x as follows and do the math:
For A, let x = -3. Does it really yield -11?
For B, let x = 0. Does it really yield 5?
For C, let x = 1/2. Does it really yield 4?
Take it from here.
The scale factor is x2 or 100. 2 times 2 is 4 and n
