A bus and a car leave from the same city traveling in opposite directions. The bus leaves 2 hours before the car does. The bus travels at a steady speed of 50 mph while the car travels at 70 mph. After how many hours will the bus and car be 800 miles apart?

Answers

Answer 1

Answer:
Normally, I would pass right by this question, because it is so complicated
and there are so many numbers in it. But I've allowed myself to become
seduced by the bounty of a full 5 points waiting at the end of the rainbow,
and I decided to rush in where more reasonable people would rightly fear
to tread.

The answer is a number, that represents "after how many hours". So we
need to know how many hours AFTER WHAT. The question is not very
clear on when the beginning of time is, so I need to decide. I'm going to
say that the time when the bus leaves is the beginning of time. And I'm
going to call ' H ' the number of hours AFTER the bus leaves.

-- The bus covers 50 miles every hour. So 'H' hours after the bus leaves,
the bus is 50H miles away from the city, in some direction.

-- The car leaves 2 hours later. So 'H' hours after the bus leaves, the car
has only been traveling for (H-2) hours.

-- The car covers 70 miles every hour. So 'H' hours after the bus leaves,
the car is (70)x(H-2) miles away from the city, in the OTHER direction.

-- Since the bus and the car are moving in opposite directions, the distance
between the bus and the car is the SUM of the distances each one covers.

   'H' hours after the bus leaves, they are (50H) plus (70)(H-2) miles apart.

What we need to figure out is: When is that distance 800 miles ?

The distance between them at 'H' hours after the bus leaves is

                                              (50H) plus (70)(H-2) miles

Eliminate the first parentheses:         50H + (70)(H-2)

Eliminate the rest of the parentheses:   50H + 70H - 140

Combine the terms with 'H' in them:         120H - 140

So here's the equation:                            120H - 140 = 800

Add 140 to each side:                            120H      = 940

Divide each side by 120 :                             H = 7.8333 hours
                                                                           = 7 and 5/6 hours
   = 7 hours and 50 minutes
   after the bus leaves.

Check:

Assume that H = 7 hrs 50 minutes (7 and 5/6 hours)
The bus has covered (50 x 7-5/6) = 391 and 2/3 miles.

The car has been traveling for only 5 and 5/6 hours
The car has covered   (70) x (5-5/6) = 408 and 1/3 miles

How far apart are they ?   (391-2/3) + (408-1/3) = 800 miles !       YAY!

Note: If the book or the homework sheet wants to count the hours
after the CAR leaves, then the answer is 5-5/6 hours, not 7-5/6 .

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Determine the exact value of k so that the quadratic function f(x) = x2 - kx + 5 has only one zero.

What is the positive solution of x2 – 36 = 5x?

Answers

Answer:

The positive solution is x=9

Step-by-step explanation:

Given the equation

$x^2-36=5x$

we have to find the positive solution of above equation.

$Equation: x^2-5x-36=0$

By splitting middle-term method

$x^2-5x-36=0$

$x^2-9x+4x-36=0$

$x(x-9)+4(x-9)=0$

$(x+4)(x-9)=0$

$x+4=0, x-9=0$

The solution is x=-4 and x=9

Hence, the positive solution is x=9

What are the values of a, b, and c in the quadratic equation –2x^2 + 4x – 3 = 0?a = 2, b = 4, c = 3

a = 2, b = 4, c = –3

a = –2, b = 4, c = 3

a = –2, b = 4, c = – 3

Answers

If you would like to find the values of a, b, and c in the quadratic equation, you can do this using the following steps:

ax^2 + bx + c = 0
-2x^2 + 4x - 3 = 0
a = -2
b = 4
c = -3

The correct result would be a = –2, b = 4, c = – 3.

Final answer:

The values of a, b, and c in the quadratic equation –2x^2 + 4x – 3 = 0 are: a = -2, b = 4, and c = -3.

Explanation:

The values of a, b, and c in the quadratic equation –2x^2 + 4x – 3 = 0 are:

a = -2

b = 4

c = -3

In a quadratic equation in the form ax^2 + bx + c = 0, the coefficients a, b, and c represent different values. In this equation, -2 is the coefficient of the x^2 term, 4 is the coefficient of the x term, and -3 is the constant term.

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Evaluate the following function for f(4).f(x) = 5 + √x

Write your answer as an integer, like this: 12

Answers

f( 4 ) = 5 + $√(4) = 5 + 2 = 7 ;$

$f(4)=5+\sqrt4=5+2=7$

Five times a number plus four times the same number is equal to 360. What is the number?

Answers

Answer:

The number is 40.

Step-by-step explanation:

Let the number be x.

Given the statement: Five times a number plus four times the same number is equal to 360.

"Five times a number" means 5x

"Four times the same number" means 4x

Then, as per the given information we have;

$5x + 4x = 360$

Combine like terms;

9x = 360

Divide both sides by 9 we get;

x = 40

Therefore, the number is, 40.

5x40=200
4x40=160
200+160=360

Find the perimeter of a rectangle whose length is 3 square root 5 and width is 2 square root 7

Answers

The rectangle has a length of 3√5 and a width of 2√7. The perimeter is calculated as 23.9 approximately.

Use the concept of a rectangle defined as:

Rectangles are four-sided polygons with all internal angles equal to 90 degrees. At each corner or vertex, two sides meet at right angles. The rectangle differs from a square in that its opposite sides are equal in length.

Given that,

Length of the rectangle: 3√5

Width of the rectangle: 2√7

To find the perimeter of a rectangle,

Use the formula P = 2(length + width).

In this case,

The length is 3√5 and the width is 2√7.

Applying the formula, we get:

P = 2(3√5 + 2√7)

To simplify this expression, distribute the 2:

P = 6√5 + 4√7

p ≈ 23.9

Hence,

The perimeter of the rectangle is approximately 23.9.

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Answer:

Step-by-step explanation:

Area=Length times width

A=3 $√(5)$ *2 $√(7)$

A=6 $√(35)$

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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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