MATHEMATICS HIGH SCHOOL

Six mobsters have arrived at the theater for the premiere of the film "Goodbuddies." One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankie’s requirement is satisfied?a. 6
b. 24
c. 120
d. 360
e. 720

Answers

Answer 1

Answer:

D = 360

Step-by-step explanation:

Number of monsters that arrive at the theater = 6

Total number of arrangement for the 6 mobsters = 6!

= 6*5*4*3*2*1

= 720

The chance that Frankie will be behind Joey is half while the chance that Joey will be behind Frankie is half.

Since Frankie wants to stand behind joey in the line though not necessarily behind him, the arrangement can be done in 6!/2 ways

= (6!) 1/2

= 720/2

= 360 ways

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Need your help! Find the length of the third side to the nearest tenth.

Answers

Answer: Im pretty sure its x = 4.472135955

Step-by-step explanation:

DISCLAIMER: I may be wrong I am kind of rusty.

You got to use the Pythagorean theorem (a^2 + b^2 = c^2) to answer this.

First you label your triangle by putting the hypotenuse as "c" the other known value as "b" and the unknown value as "a".

Now you input the answers into the equation.

x^2 + 4^2 = 6^2

Now answer the known values.

x^2 + 16 = 36

Now subtract by 16 to get rid of the sixteen.

x^2 + 16 -16 = 36 -16

x^2 = 20

Now to get rid off the ^2 find the square root.

x = 4.472135955

Write an equation in point-slope form for the line through the given point with the given slope. (9,-1); m=4/3. I need help!

Answers

$(y-y_(1) )=m(x-x_(1))$

$m= (4)/(3)$

$y_(1)=-1$

$x_(1) = 9$

$(y-(-1) )= (4)/(3) (x-(9))$

$(y+1)= (4)/(3) (x-9)$

$y - y1 = m(x - x1)$

$y - (-1) = (4)/(3) (x - 9)$

$y + 1 = (4x - 36)/(3)$

$y + 1 = (4x)/(3) - 12$

$y = (4x)/(3) - 13$

Need help with this help plz

Answers

This whole ugly thing is just 6 numbers that are all multiplied.
Here they are:

   (-7) · (x⁶) · (y⁻³) · (5) · (x⁻¹) · (y) .

To help you see what's going on, I'm going to rearrange them
and write them in a different order. (You'll remember that when
you multiply, the order of the numbers doesn't matter.)

    (-7)·(5)   ·   (x⁶)·(x⁻¹)   ·   (y⁻³)·(y) .

I'll bet you can see it already. It's really starting to fall apart.

Remember that when you have to multiply the same base to
different powers, you just add the powers.

So here's the multiplication:

   (-7)·(5)   ·   (x⁶)·(x⁻¹)   ·   (y⁻³)·(y) .
   |    |       |
      -35 · x⁵ ·       y⁻²

   -35 x⁵ y⁻²

   or you could write it as -35x⁵ / y²   which is the same thing.

Find an equation of the line that has the given slope and contains the given point. m=7/3 (1,6)

Answers

The point-slope form:

$y-y_1=m(x-x_1)$

We have the point (1, 6) and the slope m = 7/3. Substitute:

$y-6=(7)/(3)(x-1)$ use distributive property

$y-6=(7)/(3)x-(7)/(3)$ add 6 to both sides

$y=(7)/(3)x+(11)/(3)$ multiply both sides by 3

$3y=7x+11$ subtract 7x from both sides

$-7x+3y=11$ change the signs

$7x-3y=-11$

Answer:

point-slope form: $y-6=(7)/(3)(x-1)$

slope-intercept form: $y=(7)/(3)x+(11)/(3)$

standard form: $7x-3y=-11$

What are the zeros of the polynomial function f (x)=x^2-12x+20

Answers

I hope this helps you

f (x)=(x-10)(x-2)

x=10

x=2

Answer:

the answer is 10 and 2

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The volume of a frustum of a right circular cone is 52π ft3. Its altitude is 3 ft. and the measure of its lower radius is three times the measure of its upper radius. Find the lateral area of the frustum.
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In a frustum of a right circular cone, the radius of the upper base is 5 cm and the altitude is 8√3cm. If its slant height makes an angle of 60° with the lower base, find the total surface area of the frustum.
A water tank in the form of an inverted frustum of a cone has an altitude of 8 ft., and upper and lower radii of 6 ft. and 4 ft., respectively. Find the volume of the water tank and the wetted part of the tank if the depth of the water is 5 ft.
The total surface area of a frustum of a right circular cone is 435π cm2, and the base areas are 81π cm2 and 144π cm2. Find the slant height and the altitude of the frustum.
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Find the volume of a frustum of a regular square pyramid if the base edges are 13 cm and 29 cm, and the lateral edge is inclined at an angle of 45° with the lower base.
The base edges of a frustum of a regular square pyramid measure 20 cm and 60 cm. If one of the lateral edges is 75 cm, find the total surface area of the frustum.
A frustum of a regular hexagonal pyramid has an upper base edge of 16 ft. and a lower base edge of 28 ft. If the lateral area of the frustum is 1,716 ft.2, find the altitude of the frustum.
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The lateral area of a frustum of a regular triangular pyramid is 1,081 cm2, and the altitude and lateral edge are 24 cm and 26 cm, respectively. Find the lengths of the sides of the bases.

Answers

the complete answers in the attached figure

Part 1) we have

$r=4cm\n R=8 cm\n L=6cm$

Find the height h

$h^(2)=L^(2) -(R -r)^(2)\n h^(2)=6^(2) -(8-4)^(2)\n h^(2)=36-16\n h=√(20) cm$

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[8^(2) +4^(2) +8*4]√(20)\n \n V=(1)/(3)\pi[112]√(20)\n \n V=524.52 cm^(3)$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(8+4)*6\n LA=226.19 cm^(2)$

the answer Part 1) is

a) the volume is equal to $524.52 cm^(3)$

b) The Lateral area is equal to $226.19 cm^(2)$

Part 2) we have

$r=4ft\n R=5 ft\n h=100 ft$

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=100^(2) +(5-4)^(2)\n L^(2)=10000+1\n L=√(10001) ft$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(5+4)*√(10001)\n LA=2,827.57 ft^(2)$

the answer part 2) is

a) The Lateral area is equal to $2,827.57 ft^(2)$

Part 3) we have

$V=52\pi ft^(3) \n h=3ft\n R=3r$

Step 1

Find the values of R and r

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h$

substitute $R=3r$ in the formula above

$V=(1)/(3)\pi[(3r)^(2) +r^(2) +(3r)*r]*3$

$V=(1)/(3)\pi[7r)^(2)]*3$

$V=[tex] 52\pi$

$52\pi =\pi [7r^(2) ]\n r^(2) =(52)/(7) \n \n r=2.73 ft$

$R=3*2.73\n R=8.19 ft$

Step 2

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=3^(2) +(8.19-2.73)^(2)\n L^(2)=38.81\n L=6.23 ft$

Step 3

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(8.19+2.73)*6.23 LA=213.73 ft^(2)$

the answer Part 3) is

a) The lateral area is equal to $213.73 ft^(2)$

Part 4) we have

$r=15 in\n R=33 in\n h=24 in$

Find the slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=24^(2) +(33-15)^(2)\n L^(2)=576+324\n L=30 in$

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(33+15)*30\n LA=4,523.89 in^(2)$

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[33^(2) +15^(2) +33*15]24\n \n V=(1)/(3)\pi[112]24\n \n V=142.83 in^(3)$

the answer is

a) The lateral area is equal to $4,523.89 in^(2)$

b) the volume is equal to $142.83 in^(3)$

Part 5) we have

$r=5 cm\n h=8√3 cm$

Step 1

Find the value of (R-r)

$tan 60=√(3)$

$tan 60=((R-r))/(8√(3)) \n\n R-r= √(3) *8√(3) \n R-r=24 cm\n R=24+r\n R=24+5\n R=29 cm$

Step 2

Find the value of slant height L

$L^(2)=h^(2)+(R -r)^(2)\n L^(2)=(8√(3))^(2)+(24-5)^(2)\n L^(2)=192+361\n L=23.52 cm$

Step 3

Find the lateral area

$LA=\pi (R+r)L\n LA=\pi *(24+5)*23.52\n LA=2,142.82 cm^(2)$

Step 4

Find the total area

total area=lateral area+area of the top+area of the bottom

Area of the top

$r=5 cm\n A=\pi *r^(2) \n A=\pi *25\n A=78.54 cm^(2)$

Area of the bottom

$r=24 cm\n A=\pi *r^(2) \n A=\pi *576\n A=1,809.56 cm^(2)$

Total surface area

$SA=2,142.82+78.54+1,809.56\n SA=4,030.92 cm^(2)$

the answer is

a) The total surface area is $4,030.92 cm^(2)$

Part 6)

Part a) Find the volume of the water tank

we have

$r=4 ft\n R=6 ft\n h=8 ft$

Step 1

Find the volume

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[6^(2) +4^(2) +6*4]8\n \n V=(1)/(3)\pi[76]8\n \n V=636.70 ft^(3)$

the answer Part a) is $636.70 ft^(3)$

Part b) Find the volume of the wetted part of the tank if the depth of the water is 5 ft

by proportion find the radius R of the upper side for h=5 ft

$((R1-r))/(8) =((R2-r))/(5) \n\n ((6-4))/(8) =((R2-4))/(5)\n \n(R2-4)= 1.25\n R2=4+1.25\n R2=5.25 ft$

Find the volume for $R2=5.25 ft$

$V=(1)/(3)\pi[R^(2) +r^(2) +Rr]h\n\n V=(1)/(3)\pi[5.25^(2) +4^(2) +5.25*4]5\n \n V=(1)/(3)\pi[64.56]5\n \n V=338.05 ft^(3)$

the answer Part b) is $338.05 ft^(3)$

Part 7) we have

$SA=435\pi cm^(2) \n A1=144\pi cm^(2)\n A2=81\pi cm^(2)$

Step 1

Find the value of R and the value of r

$A1=\pi *R^(2) \n 144\pi =\pi *R^(2)\n R=12 cm$

$A2=\pi *r^(2) \n 81\pi =\pi *r^(2)\n r=9 cm$

Step 2

Find the value of lateral area

$LA=SA-A1-A2\n LA=435\pi -144\pi -81\pi \n LA=210\pi cm^(2)$

Step 3

Find the slant height

$LA=\pi (R+r)L\n\n L=(LA)/(\pi(R+r)) \n \n L=(210\pi)/(\pi(12+9)) \n \n L=10 cm$

Find the altitude of the frustum

$h^(2) =L^(2) -(R-r)^(2) \n h^(2) =10^(2) -(12-9)^(2)\n h^(2)=91\n h=9.54 cm$

the answer Part a) is

the slant height is $10 cm$

the answer Part b) is

the altitude of the frustum is $9.54 cm$

Find the volume and the lateral area of a frustum of a right circular cone whose radii are 4 and 8 cm, and slant height is 6 cm.
$h= √(s^2-(R_1-R_2)^2) \n = √(6^2-(4-8)^2) \n = √(36-16) \n = √(20)$
$Volume= (1)/(3) \pi h(R_1^2+R_1R_2+R_2^2) \n = (1)/(3) \pi * √(20) (4^2+4 * 8+8^2) \n = (1)/(3) \pi √(20) (16+32+64) \n = (1)/(3) \pi √(20) (112) \n =524.5cm^3$
Lateral area = Total surface area - area of base - area of top
$Lateral \ area= \pi (R_1+R_2)s \n = \pi (4+8) * 6 \n =12 \pi * 6 \n =72 \pi \n =226.2cm^2$

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