Answer:
![(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}](https://tex.z-dn.net/?f=%28-%5Cinfty%2C0%29%5Ccup%28%5Cfrac%7B16%7D%7B3%7D%2C%5Cinfty%29%5C%5CThat%20%5C%20is%20x%3C0%20%5C%20or%20%5C%20x%3E%5Cfrac%7B16%7D%7B3%7D)
Step-by-step explanation:
![h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \ respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x](https://tex.z-dn.net/?f=h%28x%29%3D%5Cfrac%7B1%7D%7B8%7Dx%5E%7B3%7D%20-x%5E%7B2%7D%20%5C%5C%5C%5CDifferentiate%20%5C%20h%28x%29%20%5C%20with%20%5C%20%20respect%20%5C%20to%20%5C%20x%5C%5Ch%27%28x%29%3D%5Cfrac%7B3%7D%7B8%7Dx%5E%7B2%7D%20-2x)
For positive rate of change ![h'(x)>0](https://tex.z-dn.net/?f=h%27%28x%29%3E0)
![\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B8%7D%20x%5E%7B2%7D%20-2x%3E0%5C%5Cx%28%5Cfrac%7B3%7D%7B8%7Dx-2%29%3E0%5C%5C%5C%5C%20When%20%5C%20x%3C0%20%5CRightarrow%20%28%5Cfrac%7B3%7D%7B8%7Dx-2%29%3C0%20%5CRightarrow%20h%27%28x%29%3E0%20%5C%20%28multiplication%20%5C%20of%20%5C%20two%20%5C%20negative%20%5C%20gives%20%5C%20positive%29%5C%5C%5C%5CWhen%20%28%5Cfrac%7B3%7D%7B8%7Dx-2%29%3E0%20%5CRightarrow%20x%3E%5Cfrac%7B16%7D%7B3%7D%20%5CRightarrow%20x%3E0%5C%5C%5C%5Ch%27%28x%29%3Dx%28%5Cfrac%7B3%7D%7B8%7Dx-2%29%20%3E0%20%5C%20%28multiplication%20%5C%20of%20%5C%20two%20%5C%20positives%20%5C%20is%20%5C%20positive%29%5C%5C%5C%5Ch%27%28x%29%3E0%20%5C%20when%20%5C%20x%3C0%20%5C%20or%20%5C%20x%3E%5Cfrac%7B16%7D%7B3%7D%20%5C%5C%5C%5C)
Y = 3x + 5 because the slope is 3/1 and 0 = 5
Yes because they are constant throughout the table.
One way to do this is to notice
![\dfrac{17\pi}{12}=\dfrac\pi6+\dfrac{5\pi}4](https://tex.z-dn.net/?f=%5Cdfrac%7B17%5Cpi%7D%7B12%7D%3D%5Cdfrac%5Cpi6%2B%5Cdfrac%7B5%5Cpi%7D4)
Then
![\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}](https://tex.z-dn.net/?f=%5Ctan%5Cdfrac%7B17%5Cpi%7D%7B12%7D%3D%5Ctan%5Cleft%28%5Cdfrac%5Cpi6%2B%5Cdfrac%7B5%5Cpi%7D4%5Cright%29%3D%5Cdfrac%7B%5Ctan%5Cfrac%5Cpi6%2B%5Ctan%5Cfrac%7B5%5Cpi%7D4%7D%7B1-%5Ctan%5Cfrac%5Cpi6%5Ctan%5Cfrac%7B5%5Cpi%7D4%7D)
We have
![\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}](https://tex.z-dn.net/?f=%5Ctan%5Cdfrac%5Cpi6%3D%5Cdfrac%7B%5Csin%5Cfrac%5Cpi6%7D%7B%5Ccos%5Cfrac%5Cpi6%7D%3D%5Cdfrac%7B%5Cfrac12%7D%7B%5Cfrac%7B%5Csqrt3%7D2%7D%3D%5Cdfrac1%7B%5Csqrt3%7D)
and since
has a period of
,
![\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1](https://tex.z-dn.net/?f=%5Ctan%5Cdfrac%7B5%5Cpi%7D4%3D%5Ctan%5Cleft%28%5Cpi%2B%5Cdfrac%5Cpi4%5Cright%29%3D%5Ctan%5Cdfrac%5Cpi4%3D1)
and so
![\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3](https://tex.z-dn.net/?f=%5Ctan%5Cdfrac%7B17%5Cpi%7D%7B12%7D%3D%5Cdfrac%7B%5Cfrac1%7B%5Csqrt3%7D%2B1%7D%7B1-%5Cfrac1%7B%5Csqrt3%7D%7D%3D%5Cdfrac%7B1%2B%5Csqrt3%7D%7B%5Csqrt3-1%7D%3D2%2B%5Csqrt3)
Answer:
x=8
Step-by-step explanation:
first you need to find the unknown corner so you do 180-71= 109
then find the interior angle degree (n-2)*180 (n=number of sides) so (4-2)*180=2(180)=360
then set everything equal to 360:
10x+6+13x-2+8x-1+109=360
(combine like terms)
31x+112=360
(subtract 112 form both sides)
31x=248
(divide both sides by 31)
x=8