Answer:
![m\angle ABC=47^o](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D47%5Eo)
Step-by-step explanation:
we know that
The measure of the external angle is the semi-difference of the arches it covers
so
![m\angle ABC=\frac{1}{2}[arc\ DE-arc\ AC]](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20DE-arc%5C%20AC%5D)
substitute the given values
![m\angle ABC=\frac{1}{2}[142^o-48^o]](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D%5Cfrac%7B1%7D%7B2%7D%5B142%5Eo-48%5Eo%5D)
![m\angle ABC=47^o](https://tex.z-dn.net/?f=m%5Cangle%20ABC%3D47%5Eo)
Answer:
His total is $269
Step-by-step explanation:
8x19 = 152
6x19.50 = 117
152+117 = 269
3) We have
![f(x) = \sec\left(\dfrac{\pi x}2\right) = \dfrac1{\cos\left(\frac{\pi x}2\right)}](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csec%5Cleft%28%5Cdfrac%7B%5Cpi%20x%7D2%5Cright%29%20%3D%20%5Cdfrac1%7B%5Ccos%5Cleft%28%5Cfrac%7B%5Cpi%20x%7D2%5Cright%29%7D)
which has vertical asymptotes (i.e. infinite discontinuities) whenever the denominator is zero. This happens for
![\cos\left(\dfrac{\pi x}2\right) = 0](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Cdfrac%7B%5Cpi%20x%7D2%5Cright%29%20%3D%200)
![\implies \dfrac{\pi x}2 = \cos^{-1}(0) + 2n\pi \text{ or } \dfrac{\pi x}2 = -\cos^{-1}(0) + 2n\pi](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B%5Cpi%20x%7D2%20%3D%20%5Ccos%5E%7B-1%7D%280%29%20%2B%202n%5Cpi%20%5Ctext%7B%20or%20%7D%20%5Cdfrac%7B%5Cpi%20x%7D2%20%3D%20-%5Ccos%5E%7B-1%7D%280%29%20%2B%202n%5Cpi)
(where
is any integer)
![\implies \dfrac{\pi x}2 = \dfrac\pi2 + 2n\pi \text{ or } \dfrac{\pi x}2 = -\dfrac\pi2 + 2n\pi](https://tex.z-dn.net/?f=%5Cimplies%20%5Cdfrac%7B%5Cpi%20x%7D2%20%3D%20%5Cdfrac%5Cpi2%20%2B%202n%5Cpi%20%5Ctext%7B%20or%20%7D%20%5Cdfrac%7B%5Cpi%20x%7D2%20%3D%20-%5Cdfrac%5Cpi2%20%2B%202n%5Cpi)
![\implies x = 1 + 4n \text{ or } x = -1 + 4n](https://tex.z-dn.net/?f=%5Cimplies%20x%20%3D%201%20%2B%204n%20%5Ctext%7B%20or%20%7D%20x%20%3D%20-1%20%2B%204n)
So the graph of
has vertical asymptotes whenever
and
.
4) Given
![h(t) = \begin{cases} t^3+1 & \text{if } t](https://tex.z-dn.net/?f=h%28t%29%20%3D%20%5Cbegin%7Bcases%7D%20t%5E3%2B1%20%26%20%5Ctext%7Bif%20%7D%20t%3C1%20%5C%5C%20%5Cfrac12%20%28t%2B1%29%20%26%20%5Ctext%7Bif%20%7D%20t%5Cge1%20%5Cend%7Bcases%7D)
we have the one-sided limits
![\displaystyle \lim_{t\to1^-} h(t) = \lim_{t\to1} (t^3+1) = 1^3+1 = 2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bt%5Cto1%5E-%7D%20h%28t%29%20%3D%20%5Clim_%7Bt%5Cto1%7D%20%28t%5E3%2B1%29%20%3D%201%5E3%2B1%20%3D%202)
and
![\displaystyle \lim_{t\to1^+} h(t) = \lim_{h\to1} \frac{t+1}2 = \frac{1+1}2 = 1](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Clim_%7Bt%5Cto1%5E%2B%7D%20h%28t%29%20%3D%20%5Clim_%7Bh%5Cto1%7D%20%5Cfrac%7Bt%2B1%7D2%20%3D%20%5Cfrac%7B1%2B1%7D2%20%3D%201)
The one-sided limits don't match, so the two-sided limit
does not exist. In other words, the limit does not exist at
because the function approaches different values from the left and right side of
.
Answer:
16
Step-by-step explanation:
The ratio fo the lengths of the sides of a 30-60-90 triangle is
short leg : long leg : hypotenuse
1 : sqrt(3) : 2
In a 30-60-90 triangle, the length of the short leg is the length of the long leg divided by sqrt(3).
The length of the short leg is 8 cm.
The length of the hypotenuse is twice the length of the short leg.
g = 2 * 8 cm = 16 cm
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