Since AE = BE = AB, \(\Delta ABE\) is an equilateral triangle

\(S_{ABE}=\dfrac{6^2\sqrt{3}}{4}=9\sqrt{3}\) (cm²)

The area of the sector ABE is : \(\dfrac{6^2\pi.60}{360}=6\pi\) (cm²)

The area of the shaded region is :

\(\dfrac{6^2\pi.90}{360}.2-\left[9\sqrt{3}+2\left(6\pi-9\sqrt{3}\right)\right]=6\pi+9\sqrt{3}\) (cm²)

The answer is : 6 + 9 + 3 = 18

We have :

\(\dfrac{1}{3}\pi R^2h=\dfrac{1}{3}\pi.3R^2.\dfrac{1}{3}h=\dfrac{1}{3}\pi\left(\sqrt{3}R\right)^2.\dfrac{1}{3}h\)

So, the radius must be increased by \(\sqrt{3}\) times or :

\(\sqrt{3}.100-100\approx73\left(\%\right)\)

\(S_{WXYZ}=XY.YZ=XY.2VY=4.\dfrac{XY.VY}{2}=4S_{XYV}=4.\dfrac{4}{5}=\dfrac{16}{5}\) (unit²)

The answer is :

\(\widehat{VEZ}=\widehat{AED}-\widehat{AEV}-\widehat{DEZ}=\dfrac{180^0.3}{5}-45^0-45^0=18^0\)

\(4x-12y=-19\Leftrightarrow12y=4x+19\Leftrightarrow y=\dfrac{1}{3}x+\dfrac{19}{12}\)

The x-coordinates of the intersection points of the graphs satisfy :

\(x^2-3x+3=\dfrac{1}{3}x+\dfrac{19}{12}\Leftrightarrow x^2-\dfrac{10}{3}x+\dfrac{17}{12}=0\)

Since there are 2 intersection points, we can apply the Vieta's formulas. The answer is \(\dfrac{10}{3}\)

When the water level rises 1 in, the amount of water in the tank increases by : 10 x 15 x 1 = 150 (in³)

The volume of each sphere is : \(\dfrac{4}{3}\pi\left(\dfrac{1}{6}:2\right)^3=\dfrac{1}{1296}\pi\) (in³)

The answer is : \(150:\dfrac{1}{1296}\pi\approx61900\) (spheres)

The sum of the factors of 120 which are different from 3 and 8 is :

4 + 5 + 6 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 + 120 = 346

So, the answer is 346

We can easily prove that EFGH is a square

\(\Rightarrow EH=EF=\dfrac{HF}{\sqrt{2}}=\dfrac{AB}{\sqrt{2}}=\sqrt{2}\left(m\right)\)

The area of the sector EFH is : \(\dfrac{\left(\sqrt{2}\right)^2\pi}{4}=\dfrac{\pi}{2}\left(m^2\right)\)

\(S_{EFH}=\dfrac{\left(\sqrt{2}\right)^2}{2}=1\left(m^2\right)\)

The answer is : \(2\left(\dfrac{\pi}{2}-1\right)=\pi-2\approx1.14\)

The volume of the cup is : \(\left(\dfrac{6}{2}\right)^2\pi.12=108\pi\) (cm³)

The volume of 48 tapioca bubbles is : \(108\pi-100\pi=8\pi\) (cm³)

The volume of each tapioca bubble is : \(\dfrac{8\pi}{48}=\dfrac{\pi}{6}\) (cm³)

The answer is : \(\sqrt[3]{\dfrac{\pi}{6}:\dfrac{4\pi}{3}}=0.5\) (cm)

The answer is : \(1029:\left(45+53\right)=10.5\) (hours)

\(BC^2=4AB.AC\Leftrightarrow AB^2+AC^2=4AB.AC\)

\(\Leftrightarrow AB^2-4AB.AC+4AC^2=3AC^2\Leftrightarrow\left(AB-2AC\right)^2=3AC^2\)

Since \(AB< AC,AB-2AC< 0\) and :

\(AB-2AC=-\sqrt{3}AC\Leftrightarrow AB=\left(2-\sqrt{3}\right)AC\)

\(\tan\widehat{C}=\dfrac{AB}{AC}=2-\sqrt{3}\Rightarrow\widehat{C}=15^0\Rightarrow\widehat{B}=75^0\)

Let v (mph), t(h), a(mi) be Bob's speed, the time Bob traveled and the answer

\(\left\{{}\begin{matrix}\left(v+5\right)\left(t+1\right)=vt+50\\\left(v+10\right)\left(t+2\right)=vt+a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}vt+v+5t+5=vt+50\\vt+2v+10t+20=vt+a\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}v+5t=45\\2v+10t=a-20\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2v+10t=90\\2v+10t=a-20\end{matrix}\right.\)\(\Rightarrow a=110\)

So, the answer is 110 miles

Let a,b be the lengths of the legs of the triangle (a > b > 0)

The hypotenuse of the triangle is : 15 x 2 = 30 (units)

\(\left\{{}\begin{matrix}ab=30.12\\a^2+b^2=30^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2-2ab+b^2=180\\a^2+2ab+b^2=1620\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=\left(6\sqrt{5}\right)^2\\\left(a+b\right)^2=\left(18\sqrt{5}\right)^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=6\sqrt{5}\\a+b=18\sqrt{5}\end{matrix}\right.\)

\(\Rightarrow2b=12\sqrt{5}\Rightarrow b=6\sqrt{5}\)

So, the answer is \(6\sqrt{5}\) units

\(\left\{{}\begin{matrix}mn=\left(m+2\right)\left(n-2\right)\\mn=\left(m-2\right)\left(n+10\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}mn-2m+2n-4=mn\\mn+10m-2n-20=mn\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2m-2n+4=0\\10m-2n-20=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}8m-24=0\\n=m+2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m=3\\n=5\end{matrix}\right.\)

The answer is : 3 x 5 = 15

Let v(mph) ,t(h) be John's velocity and the time the trip took, then vt is the distance between 2 houses

\(\left\{{}\begin{matrix}\left(v-10\right)\left(t+2\right)=vt\\\left(v+20\right)\left(t-2\right)=vt\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}vt+2v-10t-20=vt\\vt-2v+20t-40=vt\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2v-10t-20=0\\2v-20t+40=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}v=5t+10\\10t-60=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}v=40\\t=6\end{matrix}\right.\)

The answer is : 40 x 6 = 240 (mi)

In 1 minute, they can do \(\dfrac{1}{30}\) of the job together and Gertrude can do \(\dfrac{1}{40}\) of the job alone, so Bertha can do : \(\dfrac{1}{30}-\dfrac{1}{40}=\dfrac{1}{120}\) of the job alone

The answer is : 120 minutes = 2 hours

Let \(\overline{ab}\) be the number

\(\overline{ba}=\overline{ab}+\overline{ba}-\overline{ab}=10a+b+10b+a-c\left(a+b\right)\)

\(=11\left(a+b\right)-c\left(a+b\right)=\left(11-c\right)\left(a+b\right)\)

So, the answer is 11 - c

Apply the Vieta's formulas, we have : \(\left\{{}\begin{matrix}r+s=-6\\rs=-2\end{matrix}\right.\)

\(r^3+s^3=r^3+s^3+3rs\left(r+s\right)-3rs\left(r+s\right)\)

\(=\left(r+s\right)^3-3rs\left(r+s\right)=\left(-6\right)^3-3.\left(-2\right).\left(-6\right)=-252\)

\(x=\dfrac{13-73y}{2}=\dfrac{13-73.3}{2}=-103\)

\(\sqrt{4}.\dfrac{2}{3}.\sqrt{9}.\sqrt{144}\sqrt{841}=2.\dfrac{2}{3}.3.12.29=1392\)