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 : \(\left\{{}\begin{matrix}r+s=-6\\rs=-2\end{matrix}\right.\)

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

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

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

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

From 1 to 2001 :

The number of multiples of 3 is : (2001 - 3) : 3 + 1 = 667

The number of multiples of 4 is : (2000 - 4) : 4 + 1 = 500

The number of common multiples of 3 and 5 is : 

(1995 - 15) : 15 + 1 = 133

The number of common multiples of 4 and 5 is :

(2000 - 20) : 20 + 1 = 100

The number of common multiples of 3 and 4 is :

(1992 - 12) : 12 + 1 = 166

The number of common multiples of 3, 4 and 5 is :

(1980 - 60) : 60 + 1 = 33

The answer is :

(667 - 133) + (500 - 166 - 100 + 33) = 801

\(0,\overline{009}=\dfrac{1}{111}\). So, the answer is 111

The area of the patio is : 20 x 14 = 280 (ft²)

There are 2 cases (because there's no figure) :

+ The garden is tangent to 2 20-ft sides and 1 14-ft side

The radius of the garden is : \(14:2=7\left(ft\right)\)

The area of the garden is : \(7^2\pi=49\pi\left(ft^2\right)\) 

The answer is : \(\dfrac{100\left(280-49\pi\right)}{280}\%\approx45\%\)

+ The garden is tangent to 2 14-ft sides and 1 20-ft side

The radius of the garden is : \(20:2=10\left(ft\right)\)

The area of the garden is : \(10^2\pi=100\pi\left(ft^2\right)\) (asburd because \(100\pi>280\))

So, the answer is 45%

\(\left(3b^2+2b+1\right)-\left(b^2+2b+3\right)=b^2+5b+4\)

\(\Leftrightarrow b^2-5b-6=0\Leftrightarrow b^2+b-6b-6=0\)

\(\Leftrightarrow b\left(b+1\right)-6\left(b+1\right)=0\Leftrightarrow\left(b-6\right)\left(b+1\right)=0\Rightarrow b=6\)

Let a, b be the number of bicycles and tricycles in the shop

\(\left\{{}\begin{matrix}a+b=67\\2a+3b=157\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2a+2b=134\\2a+3b=157\end{matrix}\right.\)\(\Rightarrow b=23\)

So, the answer is 23

Draw \(CK\perp AB\)

\(\widehat{CBK}=180^0-\widehat{ABC}=180^0-\dfrac{180^0\left(8-2\right)}{8}=45^0\)

\(\Rightarrow CK=\dfrac{BC}{\sqrt{2}}=\sqrt{2}\)

\(S_{ABC}=\dfrac{AB.CK}{2}=\dfrac{2\sqrt{2}}{2}=\sqrt{2}\left(cm^2\right)\)

1 h = 60 m

It takes : \(60.\dfrac{300}{1200}=15\) (m) to burn 300 cal when running at 12 m/h

It takes : \(60.\dfrac{300}{450}=40\) (m) to burn 300 cal when running at 5 m/h

The answer is : 40 - 15 = 25 (m)

We notice that the sum of 2 consecutive triangular numbers is always a square number : \(\dfrac{\left(k-1\right)k}{2}+\dfrac{k\left(k+1\right)}{2}=\dfrac{k.2k}{2}=k^2\)

So, we need to find the largest triangular number less than 100

\(\dfrac{n\left(n+1\right)}{2}< 100\Leftrightarrow n^2+n< 200\Leftrightarrow n^2+2n.\dfrac{1}{2}+\dfrac{1}{4}< 200\dfrac{1}{4}\)

\(\Leftrightarrow\left(n+\dfrac{1}{2}\right)^2< 200\dfrac{1}{4}\Rightarrow n< 14\)

So, n = 13 and the answer is : \(\dfrac{13.14}{2}=91\)

(One of them is 91 and the other is \(\dfrac{12.13}{2}=78\))

The number of miles per gallon of gasoline that her old car gets is :

\(24:\left(100\%+50\%\right)=16\)

The number of gallons of gasoline that her new car spend to travel 480 miles is : 480 : 24 = 20 

The number of gallons of gasoline that her old car spend to travel 480 miles is : 480 : 16 = 30 

The answer is : 30 - 20 = 10 (gallons of gasoline)

The answer is : \(\dfrac{30}{50}+\dfrac{30}{20}=\dfrac{21}{10}=2.1\) (hours)

The statement D is true because 5 - 1 = 4

Draw \(HI,GK\perp EF\)

Each angle of the octagon is : \(\dfrac{180\left(8-2\right)}{8}=135^0\)

\(\Rightarrow\widehat{HAF}=\widehat{GFA}=180^0-135^0=45^0\)

Let a be the side length of the octagon, then \(HI=IA=GK=KF=\dfrac{a}{\sqrt{2}};IK=HG=a\)

\(\Rightarrow AF=a\left(1+\sqrt{2}\right)\Rightarrow S_{AHF}=\dfrac{a\left(1+\sqrt{2}\right).\dfrac{a}{\sqrt{2}}}{2}=\dfrac{a^2\left(2+\sqrt{2}\right)}{4}\)

\(S_{ADF}=S_{ABCDEFGH}-S_{AFGH}-S_{ABCD}-S_{FED}\)

\(=S_{AFGH}+S_{ABEF}+S_{BCDE}-S_{AFGH}-S_{ABCD}-S_{FED}\)

\(=S_{ABEF}-S_{FED}=a.a\left(1+\sqrt{2}\right)-a.\dfrac{a}{\sqrt{2}}:2\)

\(=\dfrac{a^2\left(4+3\sqrt{2}\right)}{4}\)

The answer is : \(\dfrac{4+3\sqrt{2}}{4}:\dfrac{2+\sqrt{2}}{4}=1+\sqrt{2}\approx2.41\)

Let a, b be the number of 3-ounce packages and 10-ounce packages B-Rite sold 

\(\left\{{}\begin{matrix}3a+10b< 16.16\\a+b=76\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}3a+3b+7b< 256\\3a+3b=228\end{matrix}\right.\)

\(\Rightarrow7b< 28\Rightarrow b< 4\)

So, the answer is 3

\(x-2y=8\Leftrightarrow y=\dfrac{1}{2}x-4\)

Let the equation of the line through point P be \(y=\dfrac{1}{2}x+a\left(a\ne-4\right)\)

\(3=\dfrac{1}{2}.4+a\Leftrightarrow a=1\)

So, the answer is \(y=\dfrac{1}{2}x+1\)

Let a, b be the side length of each poygon respectively \(\left(a,b\in Z^+\right)\)

\(\left\{{}\begin{matrix}6a=8b\\6a+8b< 200\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{4}{3}\\12a< 200\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a⋮4\left(b⋮3\right)\\a< 17\end{matrix}\right.\)

The answer is : (16 - 4) : 4 + 1 = 4

3 diagonals of the hexagon which are the diameters of the circle divide the hexagon into 6 congruent equilateral triangles. The area of each equilateral triangle is : \(\dfrac{\left(4\sqrt{3}\right)^2.\sqrt{3}}{4}=12\sqrt{3}\) (cm²)

The answer is : \(12\sqrt{3}.6=72\sqrt{3}\) (cm²)

Let \(\overline{ab}\) be the original integer \(\left(a,b\in N;a>0;a,b\le9\right)\)

\(\left\{{}\begin{matrix}a+b=8\\\overline{ba}=4\overline{ab}+3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=8\\a+10b=40a+4b+3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}b+a=8\\6b-39a=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2b+2a=16\\2b-13a=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}15a=15\\b=8-a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=7\end{matrix}\right.\)

So, the answer is 71

Let a be the smallest angle in degree, then the largest angle is 4a, the third angle is 2a + 10, the fourth angle is 3a - 60. We have :

\(a+4a+2a+10+3a-60=360\Leftrightarrow10a=410\Leftrightarrow a=41\Leftrightarrow4a=164\)

So, the answer is 164⁰

The number of students who participate in at least 1 activity is :

73 + 65 + (114 - 32 - 12) = 208 

The answer is : 208 + 558 = 766 (students)