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40-12=28 people had to like some flavor.

Let y be the people who like both...

Thus, we have

(18-y)+(15-y)+y=28

33-y=28

y=5

Thus, there are five(5) students in the classroom who liked both apple pie and chocolate cake.

All ten people shake hands with everyone except his/jher spouse and themseves.

Since there are 10 people(five(5) married couples), the answer should be 10 * 8  /  2  = 40 handshakes were exchanged,

Note that dividing by two cleared the overcounting of the handshakes.

The probability is 4/6=2/3  (number of blue faces)/(number of total faces).

Since the lines intersect, we can set solve for x and y using systems of equations.

3x=4y+7

y=2/5x

3x=8/5x+7

7/5x=7

x=5

y=2

Thus, (5,2)=5+2=7.

The relationship EM^2=DM*MF is very helpful...

DF=$4\sqrt{2}$, so $DM=2\sqrt{2}$ and $MF=2\sqrt{2}$.

$EM^2=2\sqrt2*2\sqrt{2}$

$EM^2=8$

Recognize that MBCN is a trapezoid-and the area can be found by: 1/2 * (sum of the bases) * (height).

The height is $\frac{40}{8}=5$ units,

Thus, the area of MBCN)(the trapezoid) is $\frac{1}{2}*6*5=15$ units squared.

By the definition of 30 - 60 - 90 triangles...AC=$2\sqrt{3}$ and $AD=\sqrt{3}$.

Triangle CDB is also a 30-60-90 degree triangle...angle B is sixty(60) degrees.

That means BD=1, and CB=2

..Thus, the area of the triangle is $\frac{1}{2}*2*2\sqrt{3}=2\sqrt{3}$

First, we solve f^-1, which can be done by setting f(y)=x...

Thus, we have 3y^2=x

$y=\frac{\sqrt{3x}}{3}$.

Thus, f^-1(12) is $y=\frac{\sqrt{3*12}}{3}=\pm2.$

The area of the triangle is 1/2*6*6.5=19.5 centimeters squared.

The area of the circle is 4^2pi=16pi.

Thu,s the probability of a point landing inside the red shaded region is 19.5/16pi$\approx$0.388

Since line A is parallel to line y=2x+4, line A must have the same slope which is two(2).

y=2x+b

Plug in (2,5) for x and y respectively

5=2(2)+b, 5=4+b, b=1

Thus, the y-intercept of line A is one(1).