What is the coefficient of x2 in the expansion of (x + 2)^3

Illusion [34]

The answer is 59/12

4 0

2 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

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The price, p, for different size orders of printed programs for a musical production, n, is given in the table.

Leya [2.2K]

The answer is B, hope this helps!

8 0

3 years ago

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Willy Wonka has 2 candies, Wonka bars and Everlasting Gobstoppers. Both have both natural sugar and sucrose in them. Each Wonka

love history [14]

Answer:

Wonka bars=3 and Everlasting Gobstoppers=24

Step-by-step explanation:

let the wonka bars be X

and everlasting gobstoppers be Y

the objective is to

maximize 1.3x+3.2y=P

subject to constraints

natural sugar

4x+2y=60------1

sucrose

x+3y=75---------2

x>0, y>0

solving 1 and 2 simultaneously we have

4x+2y=60----1

x+3y=75------2

multiply equation 2 by 4 and equation 1 by 1 to eliminate x we have

4x+2y=60

4x+12y=300

-0-10y=-240

10y=240

y=240/10

y=24

put y=24 in equation 2 we have'

x+3y=75

x+3(24)=75

x+72=75

x=75-72

x=3

put x=3 and y=24 in the objective function we have

maximize 1.3x+3.2y=P

1.3(3)+3.2(24)=P

3.9+76.8=P

80.7=P

P=$80.9

8 0

3 years ago

Solve each system using substitutions<br> Y+4=x<br> -2+ Y=8

klemol [59]

-2 + Y = 8
Y = 10
Plug in Y to the first equation, 10 + 4 = X
X = 14

3 0

3 years ago

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