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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).