To determine whether \(\displaystyle g\) is continuous at \(\displaystyle x = -\pi\), we examine the definitions of \(\displaystyle g\) on both sides of \(\displaystyle -\pi\), and evaluate both for \(\displaystyle x = -\pi\):

\(\displaystyle \sin \frac{x}{2} -1\) evaluated for \(\displaystyle x = -\pi\):

\(\displaystyle \sin \frac{-\pi}{2} -1 = \sin \left (- \frac{ \pi}{2} \right )-1= -1-1= -2\)

\(\displaystyle \cos 2x\) evaluated for \(\displaystyle x = -\pi\):

\(\displaystyle \cos 2(-\pi) = \cos (-2\pi)= 1\)

Since the values do not coincide, \(\displaystyle g\) is discontinuous at \(\displaystyle x = -\pi\).

We do the same thing with the other two boundary values 0 and \(\displaystyle \pi\).

\(\displaystyle \cos 2x\) evaluated for \(\displaystyle x =0\):

\(\displaystyle \cos 2(0) = \cos 0= 1\)

\(\displaystyle 1-\tan \frac{x }{2}\) evaluated for \(\displaystyle x =0\):

\(\displaystyle 1-\tan \frac{0 }{2} = 1 - \tan 0= 1-0 = 1\)

Since the values coincide, \(\displaystyle g\) is continuous at \(\displaystyle x =0\).

\(\displaystyle 1-\tan \frac{\pi }{2}\) turns out to be undefined for \(\displaystyle x = \pi\), (since \(\displaystyle \tan \frac{\pi }{2}\) is undefined), so \(\displaystyle g\) is discontinuous at \(\displaystyle x = \pi\).

The correct response is I and III only.

To determine whether \(\displaystyle g\) is continuous at \(\displaystyle x = -1\), we examine the definitions of \(\displaystyle g\) on both sides of \(\displaystyle -1\), and evaluate both for \(\displaystyle x = -1\):

\(\displaystyle x^{2}-2x+7\) evaluated for \(\displaystyle x = -1\):

\(\displaystyle (-1)^{2}-2(-1)+7= 1+2+7 = 10\)

\(\displaystyle 9-x\) evaluated for \(\displaystyle x = -1\):

\(\displaystyle 9-(-1)= 10\)

Since the values coincide, the graph of  \(\displaystyle g\) is continuous at \(\displaystyle x = -1\).

We do the same thing with the other two boundary values 0 and 1:

\(\displaystyle 9-x\) evaluated for \(\displaystyle x = 0\):

\(\displaystyle 9-0= 9\)

\(\displaystyle x+8\) evaluated for \(\displaystyle x = 0\):

\(\displaystyle 0+8 = 8\)

Since the values do not coincide, the graph of \(\displaystyle g\) is discontinuous at \(\displaystyle x = 0\).

\(\displaystyle x+8\) evaluated for \(\displaystyle x = 1\):

\(\displaystyle 1+8 = 9\)

\(\displaystyle x^{2}+3x+ 6\) evaluate for \(\displaystyle x = 1\):

\(\displaystyle 1^{2}+3\cdot 1+ 6 = 1+ 3 + 6 = 10\)

Since the values do not coincide, the graph of \(\displaystyle g\) is discontinuous at \(\displaystyle x = 1\).

II and III only is the correct response.

To find the \(\displaystyle y\)-intercept, evaluate \(\displaystyle g(0)\) using the definition of \(\displaystyle g\) on the interval that includes the value 0. Since 

\(\displaystyle g(x)= \cos 2x\)

on the interval  \(\displaystyle (-\pi, 0]\),

evaluate:

\(\displaystyle g(x)= \cos 2 \cdot 0 = \cos 0 = 1\)

The \(\displaystyle y\)-intercept is \(\displaystyle (0, 1)\).

The period of the graph of a cosine function \(\displaystyle f(x)= A \cos Nx\) is \(\displaystyle \frac{2 \pi}{N}\), or \(\displaystyle 2 \pi \div N\)

Since  \(\displaystyle N = \frac{3\pi}{7}\), the period is

\(\displaystyle 2 \pi \div \frac{3\pi}{7} = \frac{2 \pi }{1}\times \frac{7} {3\pi}= \frac{14}{3}\)

\(\displaystyle BD\), as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

\(\displaystyle BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12\).

The area of \(\displaystyle \Delta ADB\), the shaded region, is half the products of its legs:

\(\displaystyle A= \frac{1}{2} (AD) \times (DB ) =\frac{1}{2} \times 6 \times 12 = 36\)

The area of \(\displaystyle \Delta ABC\) is half the product of its hypoteuse, which we can see as the base, and the length of corresponding altitude  \(\displaystyle \overline{BD}\):

\(\displaystyle A = \frac{1}{2} (AC) (BD) = \frac{1}{2} \cdot 30 \cdot 12 = 180\)

\(\displaystyle \Delta ADB\) comprises

\(\displaystyle \frac{36}{180} \times \100 = 20 \%\)

of \(\displaystyle \Delta ADC\).

The area of the dirt path is the area of the outer square minus that of the inner square.

The outer square has sidelength 75 feet and therefore has area

\(\displaystyle 75 \times 75 = 5,625\) square feet.

The inner square has sidelength \(\displaystyle 75 - 2 \times 7 = 61\) feet and therefore has area 

\(\displaystyle 61 \times 61 = 3,721\) square feet.

Subtract to get the area of the dirt path:

\(\displaystyle 5,625 - 3,721 = 1,904\) square feet.

The length of the garden is \(\displaystyle 2 \times 6= 12\) feet less than that of the entire lot, or 

\(\displaystyle L = 2x-12\);

The width of the garden is \(\displaystyle 2 \times 6= 12\) less than that of the entire lot, or 

\(\displaystyle W = x-12\);

The area of the garden is their product:

\(\displaystyle A = LW = (2x-12)(x - 12)\)

\(\displaystyle =2x \cdot x - 2x \cdot 12 - 12 \cdot x + 12 \cdot 12\)

\(\displaystyle = 2x ^{2} - 24x - 12 x + 144\)

\(\displaystyle = 2x ^{2} - 36 x + 144\)

As an interior angle of a regular decagon, \(\displaystyle \angle B\) measures

\(\displaystyle m \angle B = \left [\frac{180(10-2)}{10} \right ] ^{\circ }= 144^{\circ }\).

\(\displaystyle AB = BC = 100\).

\(\displaystyle AC\) can be found using the Law of Cosines:

\(\displaystyle \left ( AC\right )^{2} = \left ( AB\right )^{2}+ \left ( BC\right )^{2} - 2 \left ( AB\right ) \left ( BC\right ) \cos m\angle B\)

\(\displaystyle \left ( AC\right )^{2} = 100^{2}+ 100^{2} - 2 \cdot 100 \cdot 100 \cos 144^{\circ }\)

\(\displaystyle \left ( AC\right )^{2} \approx 10,000 + 10,000 - 20,000 (-0.8090)\)

\(\displaystyle \left ( AC\right )^{2} \approx 20,000 - (-16,180) \approx 36,180\)

\(\displaystyle AC \approx \sqrt{36,180} \approx 190\)

First, it is necessary to determine the radius of the circle. This is the distance between \(\displaystyle (0,0)\) and \(\displaystyle (-8,6)\), so we apply the distance formula:

\(\displaystyle r = \sqrt{(x_{2} - x _{1})^{2}+(y_{2} - y_{1})^{2}}\)

\(\displaystyle r = \sqrt{\left ( -8 -0 \right )^{2} +\left ( 6-0\right )^{2} }=\sqrt{ 8^{2}+6^{2}} = \sqrt{64+36 }= \sqrt{100} = 10\)

Subsequently, the area of the circle is 

\(\displaystyle A = \pi r^{2} = \pi \cdot 10^{2} = 100 \pi\)

Now, we need to find the central angle of the shaded sector. This is found using the relationship

\(\displaystyle \tan \theta = \frac{y}{x}\)

\(\displaystyle \tan \theta = \frac{6}{-8} = -0.75\)

Using a calculator, we find that \(\displaystyle \tan^{-1}(-0.75) \approx -36.87^{\circ }\); since we want a degree measure between \(\displaystyle 90^{\circ }\) and \(\displaystyle 180 ^{\circ }\), we adjust by adding \(\displaystyle 180 ^{\circ }\), so

\(\displaystyle \theta \approx \left (-36.87+ 180 \right ) ^{\circ } \approx 143.13^{\circ }\)

The area of the sector is calculated as follows:

\(\displaystyle A \approx \frac{143.13 }{360} \cdot 100 \pi \approx 124.9\)

You own a mug with a circular bottom. If the distance around the outside of the mug's base is  \(\displaystyle 20 \pi cm\) what is the area of the base?

Begin by solving for the radius:

\(\displaystyle C=2\pi r\)

\(\displaystyle 20 \pi cm=2\pi r\)

\(\displaystyle r=10cm\)

Next, plug the radius back into the area formula and solve:

\(\displaystyle A_{circle}=\pi r^2\)

\(\displaystyle A_{circle}=\pi (10cm)^2=100 \pi cm^2\)

So our answer is:

\(\displaystyle 100 \pi cm^2\)