Answer:
145
Step-by-step explanation:
angles in a pentagon add up to equal 540
Hence, 110 + 90 + 80 + 115 + v = 540
( note that we just created an equation that we can use to solve for v )
We now solve for v using the equation created
110 + 90 + 80 + 115 + v = 540
step 1 combine like terms
110 + 90 + 80 + 115 = 395
we now have 395 + v = 540
step 2 subtract 395 from each side
540 - 395 = 145
395 - 395 cancels out
we're left with v = 145
Answer:
The estimated sum is ≈10
Step-by-step explanation:
This depends on how your professor wants you to estimate sums. Logically, though, a value of nearly 4 and a value of just over 6 add to about 10. The exact answer would be a bit over 10, at ≈10.14117
T = days passed
r = rate of growth
by 0 day, or t = 0, there are 2 folks sick,

by the third day, t = 3, there are 40 folks sick,
![\bf \qquad \textit{Amount for Exponential Growth} \\\\ A=P(1 + r)^t\qquad \begin{cases} A=\textit{accumulated amount}\to &40\\ P=\textit{initial amount}\to &2\\ r=rate\to r\%\to \frac{r}{100}\\ t=\textit{elapsed time}\to &3\\ \end{cases} \\\\\\ 40=2(1+r)^3\implies 20=(1+r)^3\implies \sqrt[3]{20}=1+r \\\\\\ \sqrt[3]{20}-1=r\implies 1.7\approx r\qquad \boxed{A=2(2.7)^t}](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7BAmount%20for%20Exponential%20Growth%7D%0A%5C%5C%5C%5C%0AA%3DP%281%20%2B%20r%29%5Et%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0AA%3D%5Ctextit%7Baccumulated%20amount%7D%5Cto%20%2640%5C%5C%0AP%3D%5Ctextit%7Binitial%20amount%7D%5Cto%20%262%5C%5C%0Ar%3Drate%5Cto%20r%5C%25%5Cto%20%5Cfrac%7Br%7D%7B100%7D%5C%5C%0At%3D%5Ctextit%7Belapsed%20time%7D%5Cto%20%263%5C%5C%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0A40%3D2%281%2Br%29%5E3%5Cimplies%2020%3D%281%2Br%29%5E3%5Cimplies%20%5Csqrt%5B3%5D%7B20%7D%3D1%2Br%0A%5C%5C%5C%5C%5C%5C%0A%5Csqrt%5B3%5D%7B20%7D-1%3Dr%5Cimplies%201.7%5Capprox%20r%5Cqquad%20%5Cboxed%7BA%3D2%282.7%29%5Et%7D)
how many folks are there sick by t = 6?
Let ∠RTS=∠RST = a (say)
∠QUA=∠QSU= b(say)
then we know , at point S, a+40+b=180. so, a+b=140 we'll use this later.
consider trianglePQR, ∠P+∠Q+∠R=180
i.e.P+(180-2b)+(180-2a)=180
P+180+180-2(a+b)=180 ⇒P+180-2(a+b)=0 ⇒P=2(140)- 180=280-180=100
hence,answer is E