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The  circumference  of the cone =  2 * pi * 21  * (120/360)   =  14 pi

The radius of the cone = 

14pi =  2 * pi * r

7 = r

Height of the cone  = sqrt [ 21^2 - 7^2 ]  = sqrt (392) = 14sqrt 2

Volume of cone =   (1/3) pi (7)^2  * 14sqrt 2   = 686sqrt (2 ) / 3   pi   cm^3

Formula  for the number of lines, n,  that will divide a plane into k regions

(1/2) (n^2 + n + 2)  =  k

(1/2) (n^2 + n + 2)  =   5

n^2 + n + 2  = 10

n^2 + n =  8

n  must be  greater than 2

Here : https://web2.0calc.com/questions/sphere_18

Nothing shown

Sorry....I thought this problem was similar to another one asked on here earlier.....see my corrected answer ....

What is the difference between the surface area of the sphere and the lateral surface area of the cylinder?

Hello tomtom!

\(V_c =\pi r^2\cdot 2r=2\pi r^3=144\\ r=(\dfrac{72}{\pi })^{\frac{1}{3}}\)

\(LS_c=2\pi r^2+2\pi r\cdot 2r=2\pi r^2(1+2)=6\pi r^2\\ SA_s=4\pi r^2\\ LS_c-SA_s= 2\pi r^2=2\pi \cdot (\dfrac{72}{\pi })^{\frac{2}{3}} \\ \color{blue}LS_c-SA_s=50.6954\)

The surface area of the sphere and the lateral surface area of the cylinder is 50.6954.

 !

The problem says round trip, from Penthaven to Jackson and back. Then the problem says that the wind is blowing from Penthaven to Jackson during our round trip.

5 hrs 20 min =  5 + 1/3 hrs =  (16/3)hrs

Call the rate of the plane in still air =  R

With the wind the rate is R + 75

Against the wnid the rate is R - 75 

Call the distance between the cities D

D/R = T

In the first instance

Total Distance / Rate in still air = 16/3 hrs

2D / R   =  (16/3)

D/R = 8/3

D = (8/3)R      (1)

In the second instance

Time from Penthaven to Jackson  + Time from Jackson to Penthaven  =  6  hrs

D / ( R + 75 )  +  D / ( R - 75)  =  6      (2)

Sub (1)  into (2)

(8/3)R / ( R + 75)  + (8/3)R / ( R -75)   = 6

(8/3)R ( R -75)  + (8/3)R (R + 75)  =  6 ( R+ 75)(R -75)

(8/3)R^2 - 200R + (8/3)R^2 + 200R  =  6 ( R^2 - 5625)

(16/3)R^2  = 6R^2 - 33750

R^2 ( 6 -16/3)  = 33750

(2/3)R^2 = 33750

R^2  = 50625

R=  225

The distance between the  cities 

D = (8/3)225 =   600  miles 

f (f ( f(x)))  =  66x - 47

f (f (x) ) =  p(px + q) + q =  p^2x + pq + q

f ( f ( f (x)) )  =  p (p^2x + pq + q) + q =   p^3x + p^2q + pq + q

So

p^3x + p^2q + pq + q  =  66x - 47

p^3  = 66

p= 66^(1/3)

And

p^2q + pq + q  =  -47

66^(2/3)q + 66^(1/3)*q + q = -47

q (66^(2/3) + 66^(1/3) + 1)  =  -47

q  = -47 / ( 66^(2/3) + 66^(1/3) + 1)

p + q =     [ 66^(1/3) ( 66^(2/3) + 66^(1/3) + 1)  - 47] / [ 66^(2/3) +66^(1/3) + 1 ]  =

[ 66^(2/3) + 66^(1/3) + 19 ]  /  [66^(2/3) + 66^(1/3) + 1 ]  ≈  1.842..... 

The x coordinate of the vertex =  -1  =    -b / (2a)

So

-b / ( 2 * 2)   = -1

-b / 4 =  -1

b/4 = 1

b  = 4

Simplify as

y = -5x^2 + 15x - 10

x coordinate of the  vertex  = (-15) / (2 * -5)  = 3/2

y coordinate of the  vertex  = 

 -5(3/2)^2  + 15(3/2) - 10  =  

-45/4 + 45/2 - 10 =

45/4 -10  =

5/4

Vertex = ( 3/2, 5/4)  = ( 1.5 1.25)

\( f(x) + b \cdot g(x)\)

If b = 0, the resulting polynomial will be degree 5

No triangle can be constructed with the given data.

AB must be greater than BC.

  !

Find the smallest angle, in degrees.

Hello blackpanther!

\(\alpha +\alpha +d +\alpha +2d=180^\circ\\ (\alpha +2d)-\alpha =56^\circ\\ d=28^\circ\\ 3\alpha +3\cdot 28^\circ =180^\circ\\ \alpha+28^\circ =60^\circ\\ \color{blue}\alpha =32^\circ\\ \beta =60^\circ\\ \gamma =88^\circ \)

\(\color{blue}\alpha ,the\ smallest\ angle\ is\ 32^\circ .\) 

 !

There cannot be more than 3 yelllow squares

Four ways

Y B Y R Y

Y R Y B Y

Y B Y B Y

Y R Y R Y

      1     =     row 0

   1    1

 1   2    1

1  3    3    1

The sum of the entries in the nth row of Pascal's Triangle  =  2^n

The number of entries in any row  =  n + 1     

So  the average is

[ 2^n ] / [ n + 1]  = 2

2^n  = 2( n + 1)

2^n =  2n + 2

Note that  this is  true if n = 3

Angle AED = 90  = Angle BED

DC = DE

DB  = DB

Triangle DBC congrunent to triangle DBE  by HL

tan (DBC)  = DC / BC  = (BC/3) / BC   = 1/3  = tan (DBE)

tan (EBC)  =  AC  / BC  = 

tan (2 * DBC)   = 

 [ tan (DBE) + tan (DBC) ]  / [ 1 - tan(DBE)*tan (DBC) ] =

(1/3 + 1/3 )  /  ( 1 - (1/3)(1/3) )  =

(2/3) / ( 8/9)  =

18/24 =

3/4 =  AC / BC

\(\dfrac{s}{25\ miles/h}-\dfrac{s}{40\ miles/h}=\dfrac{45}{60}h\\ \dfrac{s(40-25)\ miles/h}{1000\ (miles/h)^2}=\dfrac{45}{60}h\\ \dfrac{15s}{1000\ miles/h}=\dfrac{45}{60}h\\ s=\dfrac{45h\cdot1000\ miles/h}{60\cdot15}\\ \color{blue}s=50\ miles\)

  !

how many ways to make a dollar using 4 quarters 6 dimes and 8 nickels?   
The answer I got was 13 and I was just looking for confirmation.  

4 Quarters                                                       1 way  

                     4 quarters  

3 Quarters                                                        3 ways  

                     3 quarters, 2 dimes, 1 nickel 

                     3 quarters, 1 dime, 3 nickels 

                     3 quarters, 0 dime, 5 nickels  

2 Quarters                                                         5 ways  

                     2 quarters, 5 dimes, 0 nickel  

                     2 quarters, 4 dimes, 2 nickels  

                     2 quarters, 3 dimes, 4 nickels  

                     2 quarters, 2 dimes, 6 nickels  

                     2 quarters, 1 dime, 8 nickels  

1 Quarter                                                            3 ways  

                     1 quarter, 6 dimes, 3 nickels  

                     1 quarter, 5 dimes, 5 nickels  

                     1 quarter, 4 dimes, 7 nickels  

0 Quarter                                                             1 way  

                     0 quarter, 6 dimes, 8 nickels  

                                                           total          13 ways  

You found 13 ways.  I can't find a 14th one, 

so it looks like we both got the same answer. 

_.

Hello tomtom, hello greenplanet2050!

The left function is an upwardly open second power parabola above the x-axis,

the left function is a downwardly open second power parabola under the x-axis.
For every x value there is no real function value that is included in both functions at the same time.

The graphs do not touch each other.

  !

Actually, I solved it! If anyone needs help? Here a hint: Use difference of squares: \(x^2-y^2=(x+y)(x-y)\) and prime factorization! 

Let (x,y) be a point on the graph of y=1/2​(x2−18). The distance between this point and the origin is given by the distance formula:

x2+y2​=x2+(1/2​(x2−18))2​.

Squaring both sides, we get:

x2+21​(x2−18)2=x4−4x2+81=(x2−2)2+77.

To minimize the distance, we want to minimize x2−2. Since the square of a real number is non-negative, its minimum value is 0. Therefore, the minimum value of x2−2 is 0, which occurs when x=±2. Substituting x=±2 back into the equation for the distance, we get:

sqrt(2^2 + (1/sqrt(2)((2)^2 - 18))^2) = sqrt(4 + 1/2) = sqrt(9/2) = 3/sqrt(2) = 3*sqrt(2)/2

Therefore, the answer is 3*sqrt(2)/2.

We know that since f(x) is divisible by x3, it can be written in the form f(x)=ax5+bx4+cx3+dx2+ex+f.

We are also given that f(x)+3 is divisible by (x+1)3, which can be expanded as (x+1)(x+1)(x+1)=x3+3x2+3x+1.

This tells us that there exist polynomials q(x) and r(x) such that:

f(x)+3=(x3+3x2+3x+1)q(x)+r(x)

Since (x3+3x2+3x+1) has a degree of 3, and we are given that f(x)+3 is divisible by it, then the remainder r(x) must be a polynomial of degree at most 2

(because any higher degree term in r(x) would be canceled out by the division). This means r(x)=mx2+nx+p for some constants m, n, and p.

Substituting the form of f(x) and expanding the equation, we get:

ax5+bx4+cx3+dx2+ex+f+3=(x3+3x2+3x+1)q(x)+mx2+nx+p

Matching coefficients of like terms on both sides, we get the following system of equations:

a=0 (since the coefficient of x5 on the right side is 0)

b=m+3q(1)

c=n+3q(0)+1 (since the constant term on the right side is 1)

d=p+3q(−1)

e=3q(−2)

f=p+3q(−3)

Since we are looking for the smallest possible value of f, we want to minimize the constant term p. From equation (6), we see that f is minimized when p is

minimized.

Setting p=0 and solving the remaining equations, we get:

a=0

b=m+3

c=n+1

d=−3

e=−6

f=−9

Therefore, the polynomial f(x) that satisfies the given conditions is:

f(x)=4x5+(m+3)x4+(n+1)x3−3x2−6x−9

The final step is to find the values of m and n that minimize f(x). Since we can choose any values for m and n as long as p=0, we can simply set them to 0.

This results in the polynomial:

f(x) = 4x^5 + 3x^4 + x^3