Find the distance traveled by a particle with position (x, y) as t varies in the given time interval.
x = 3 sin^2(t), y = 3 cos^2(t), 0< t<3pi
What is the length of the curve?

Find The Distance Traveled By A Particle With Position (x, Y) As T Varies In The Given Time Interval.x

Answers

Answer 1

The length of the curve (and thus the total distance traveled by the particle along the curve) is

[tex]\displaystyle\int_0^{3\pi}\sqrt{x'(t)^2+y'(t)^2}\,\mathrm dt[/tex]

We have

x(t) = 3 sin²(t )   ==>   x'(t) = 6 sin(t ) cos(t ) = 3 sin(2t )

y(t) = 3 cos²(t )   ==>   y'(t) = -6 cos(t ) sin(t ) = -3 sin(2t )

Then

√(x'(t) ² + y'(t) ²) = √(18 sin²(2t )) = 18 |sin(2t )|

and the arc length is

[tex]\displaystyle 18 \int_0^{3\pi} |\sin(2t)| \,\mathrm dt[/tex]

Recall the definition of absolute value: |x| = x if x ≥ 0, and |x| = -x if x < 0.

Now,

• sin(2t ) ≥ 0 for t ∈ (0, π/2) U (π, 3π/2) U (2π, 5π/2)

• sin(2t ) < 0 for t ∈ (π/2, π) U (3π/2, 2π) U (5π/2, 3π)

so we split up the integral as

[tex]\displaystyle 18 \left(\int_0^{\pi/2} \sin(2t) \,\mathrm dt - \int_{\pi/2}^\pi \sin(2t) \,\mathrm dt + \cdots - \int_{5\pi/2}^{3\pi} \sin(2t) \,\mathrm dt\right)[/tex]

which evaluates to 18 × (1 - (-1) + 1 - (-1) + 1 - (-1)) = 18 × 6 = 108.


Related Questions


Which is the solution to-x/2<-4
A x<-8
B x2-8
C x <8
D x 8

Answers

Answer:

A.x<-8

Step-by-step explanation:

=1/2x<−4

=2*(1/2x)< (2)*(-4)

= x<-8

Hector's Position:
Hector was standing halfway between first and second base, at the grass line. The
grass line is 95 feet from the pitcher's mound.
6. Calculate the coordinates for Hector's position. [Note: We can assume that 95
feet is an approximately horizontal distance from the pitcher's mound to the grass
line.] (2 points: 1 for x, 1 for y)
Hector was standing at the coordinate ( __, _).
Calculate Hector's Throw:

Answers

Answer:

(137.78, 47.72)

Step-by-step explanation:

(I just finished this assignment.)

Tre's position at the pitcher's mound as the point (42.78, 42.78).

                                                                                  (     x    ,    y     )

Hector is about 95 feet away from the pitcher's mound horizontal, (x axis).

Since we already have the correct y-coordinate, we need to solve for the correct x-coordinate.

  x = 95 + 42.78

         ↓ ↓ ↓

95 + 42.78 = 137.72

Now all you need to do is write out the coordinates.

Hector's coordinates are (137.72, 47.78 )

1+9x=80 Find for x
Kid can’t figure it out and I don’t know this stuff

Answers

Answer:

79/9  or 8 7/9

Step-by-step explanation:

1+9x=80

1 +9x = 90

Subtract 1 from each side

1+9x-1 = 80-1

9x = 79

divide by 9

9x/9 = 79/9

x = 8 7/9

Answer:

Step-by-step explanation:

1+9x = 80

9x = 80-1

9x = 79

x = 79/9

or

x = 8 7/9

or

x ≈ 8.78 (x = about 8.78) rounded

f(t)= 102,000/1+4400e^-t

Answers

Answer:

Beginning (t=0) population with flu is 23.

After 4 weeks, population with flu is 1250.

After an infinite amount of weeks, the population witf flu is 102000

Step-by-step explanation:

First question asks you to replace t with 0 because it says beginning.

102000/(1+4400e^-0)=102000/(1+4400)=102000/4401=23.17655 approximately. To nearest whole number this is 23.

After 4 weeks means we replace t with 4:

102000/(1+4400e^-4)

Calculator time:

1250.17142 which to nearest whole number is 1250

If t is super large, then e^-t is super close to 0.

So the limiting number is

102000/(1+4400×0)=102000/1=102000

Charlie has an annual salary of $75,000.00. He is paid every two weeks. What is the gross income amount for each paycheck?

Answers

Answer:

$2884.62

Step-by-step explanation:

A year has 52 weeks

The number of times Charlie will receive a paycheck will be 52w ÷ 2w = 26 times

Charlie's gross income each paycheck will be 7500÷26 = $2884.62 every two weeks

75000 ÷ (52 ÷2)

7500 ÷ 26

$2884.62

write the expression as a decimal , 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000 =__

Answers

Answer:

6.986.

Step-by-step explanation:

6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

We do the multiplications first    ( according to PEMDAS):-

= 6 + 9 * 0.1 + 8 * 0.01 + 6 * 0.001

= 6 + 0.9 + 0.08 + 0006

= 6.9 + 0.086

= 6 986.

The value of the equation in the decimal form is A = 6.986

What is an Equation?

Equations are mathematical statements with two algebraic expressions flanking the equals (=) sign on either side.

It demonstrates the equality of the relationship between the expressions printed on the left and right sides.

Coefficients, variables, operators, constants, terms, expressions, and the equal to sign are some of the components of an equation. The "=" sign and terms on both sides must always be present when writing an equation.

Given data ,

Let the equation be represented as A

Now , the value of A is

A = 6 x 1 + 9 x 1/10 + 8 x 1/100 + 6 x 1/1000

On simplifying the equation , we get

The value of 6 x 1 = 6

The value of 9 x 1/10 = 0.9

The value of 9 x 1/100 = 0.08

The value of 6 x 1/1000 = 0.006

So , substituting the values in the equation A , we get

A = 6 + 0.9 + 0.08 + 0.006

On simplifying the equation , we get

A = 6.986

Therefore , the value of A is 6.986

Hence , the value of the equation is 6.986

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To calculate the volume of a chemical produced in a day a chemical manufacturing company uses the following formula below:
[tex]V(x)=[C_1(x)+C_2(x)](H(x))[/tex]
where represents the number of units produced. This means two chemicals are added together to make a new chemical and the resulting chemical is multiplied by the expression for the holding container with respect to the number of units produced. The equations for the two chemicals added together with respect to the number of unit produced are given below:
[tex]C_1(x)=\frac{x}{x+1} , C_2(x)=\frac{2}{x-3}[/tex]
The equation for the holding container with respect to the number of unit produced is given below:
[tex]H(x)=\frac{x^3-9x}{x}[/tex]

a. What rational expression do you get when you combine the two chemicals?
b. What is the simplified equation of ?
c. What would the volume be if 50, 100, or 1000 units are produced in a day?
d. The company needs a volume of 3000 How many units would need to be produced in a day?

Answers

Answer:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]V(50) = 2548.17[/tex]        [tex]V(100) = 10098.10[/tex]       [tex]V(1000) = 999201.78[/tex]

[tex]x = 54.78[/tex]

Step-by-step explanation:

Given

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

[tex]C_1(x) = \frac{x}{x+1}[/tex]

[tex]C_1(x) = \frac{2}{x-3}[/tex]

[tex]H(x) = \frac{x^3 - 9x}{x}[/tex]

Solving (a): Expression for V(x)

We have:

[tex]V(x) = [C_1(x) + C_2(x)](H(x))[/tex]

Substitute known values

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solving (b): Simplify V(x)

We have:

[tex]V(x) = [\frac{x}{x + 1} + \frac{2}{x-3}] * \frac{x^3 - 9x}{x}[/tex]

Solve the expression in bracket

[tex]V(x) = [\frac{x*(x-3) + 2*(x+1)}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-3x + 2x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x^3 - 9x}{x}[/tex]

Factor out x

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * \frac{x(x^2 - 9)}{x}[/tex]

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x^2 - 9)[/tex]

Express as difference of two squares

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)(x -3)}] * (x- 3)(x + 3)[/tex]

Cancel out x - 3

[tex]V(x) = [\frac{x^2-x+2}{(x + 1)}] *(x + 3)[/tex]

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Solving (c): V(50), V(100), V(1000)

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Substitute 50 for x

[tex]V(50) = [\frac{(50^2-50+2)(50 + 3)}{(50 + 1)}][/tex]

[tex]V(50) = \frac{(2452)(53)}{(51)}][/tex]

[tex]V(50) = 2548.17[/tex]

Substitute 100 for x

[tex]V(100) = [\frac{(100^2-100+2)(100 + 3)}{(100 + 1)}][/tex]

[tex]V(100) = \frac{9902)(103)}{(101)}[/tex]

[tex]V(100) = 10098.10[/tex]

Substitute 1000 for x

[tex]V(1000) = [\frac{(1000^2-1000+2)(1000 + 3)}{(1000 + 1)}][/tex]

[tex]V(1000) = [\frac{(999002)(10003)}{(10001)}][/tex]

[tex]V(1000) = 999201.78[/tex]

Solving (d): V(x) = 3000, find x

[tex]V(x) = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

[tex]3000 = [\frac{(x^2-x+2)(x + 3)}{(x + 1)}][/tex]

Cross multiply

[tex]3000(x + 1) = (x^2-x+2)(x + 3)[/tex]

Equate to 0

[tex](x^2-x+2)(x + 3)-3000(x + 1)=0[/tex]

Open brackets

[tex]x^3 - x^2 + 2x + 3x^2 - 3x + 6 - 3000x - 3000 = 0[/tex]

Collect like terms

[tex]x^3 + 3x^2- x^2 + 2x - 3x - 3000x + 6 - 3000 = 0[/tex]

[tex]x^3 + x^2 -3001x -2994 = 0[/tex]

Solve using graphs (see attachment)

[tex]x = -54.783[/tex] or

[tex]x = -0.998[/tex] or

[tex]x = 54.78[/tex]

x can't be negative. So:

[tex]x = 54.78[/tex]

Beginning in January, a person plans to deposit $1 at the end of each month into an account earning
15% compounded monthly. Each year taxes must be paid on the interest earned during that year. Find
the interest earned during each year for the first 3 years.

Answers

Answer:

hi I am a Nepal

[tex] {233333}^{2332} [/tex]

240) What term is 359 in the sequence 5, 12, 17, 23, 28, 29......?​

Answers

Answer:

Check your question again

Step-by-step explanation:

The arithmetic equation of this sequence is an=5+(n-1)*7. Replace 359 with an and solve for n

359=5+(n-1)*7, 354/7=n-1. Wait you got the whole equation wrong, the first term should be 7 so that the common difference be equal to 5

Plz urgennt look at the image over 1000 points im going to need help with the last 4 questions i have?

Answers

The answer should be 1548.3m²

A triangular lamina has vertices (0, 0), (0, 1) and (c, 0) for some positive constant c. Assuming constant mass density, show that the y-coordinate of the center of mass of the lamina is independent of the constant c.

Answers

The equation of the line through (0, 1) and (c, 0) is

y - 0 = (0 - 1)/(c - 0) (x - c)   ==>   y = 1 - x/c

Let L denote the given lamina,

L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}

Then the center of mass of L is the point [tex](\bar x,\bar y)[/tex] with coordinates given by

[tex]\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m[/tex]

where [tex]M_x[/tex] is the first moment of L about the x-axis, [tex]M_y[/tex] is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.

Let ρ be the mass density of L. Then L has a mass of

[tex]\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2[/tex]

Now we compute the first moment about the y-axis:

[tex]\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6[/tex]

Then

[tex]\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3[/tex]

but this clearly isn't independent of c ...

Maybe the x-coordinate was intended? Because we would have had

[tex]\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6[/tex]

and we get

[tex]\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13[/tex]

The center of mass for a uniform triangular shape is on its centroid. The y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

What is the center of mass for a triangular shape?

If the surface is plane triangle approximately and mass is uniformally distributed, then its center of mass will lie on the centroid of that triangle.

What is centroid of a triangle and its coordinates?

The point of intersection of a triangle's medians is its centroid (the lines joining each vertex with the midpoint of the opposite side).

If the triangle has its vertices as  [tex](x_1, y_1), (x_2, y_2) , \: (x_3, y_3)[/tex], then the coordinates of the centroid of that triangle is given by:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)[/tex]

For this case, the  triangular lamina has vertices (0, 0), (0, 1) and (c, 0)

Assuming its mass is spread regularly, the coordinates of its center of mass would be:

[tex](x,y) = \left( \dfrac{x_1 + x_2 + x_3}{3} + \dfrac{y_1 + y_2 + y_3}{3} \right)\\\\(x,y) = \left( \dfrac{0+0+c}{3} + \dfrac{0+1+0}{3} \right) = (c/3, 1/3)[/tex]

Thus, the y-coordinate of the center of mass of the lamina is 1/3 (independent of c).

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At the movie theatre, child admission is $5.80 and adult admission is $9.70. On Wednesday, 171 tickets were sold for a total sales of $1296.00. How many
child tickets were sold that day?

Answers

Answer:

93 child tickets

Step-by-step explanation:

Create a system of equations where c is the number of child tickets sold and a is the number of adult tickets sold:

c + a = 171

5.80c + 9.70a = 1296.00

Solve by elimination by multiplying the top equation by -9.7:

-9.7c - 9.7a = -1658.7

5.8c + 9.7a = 1296

Add these together and solve for c:

-3.9c = -362.7

c = 93

So, 93 child tickets were sold.

A ball is dropped from a height of 14 ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen. (a) Find the total distance the ball has traveled at the instant it hits the ground the fourth time. (Enter an exact number.)

Answers

Answer:

Hello,

742/27 (ft)

Step-by-step explanation:

[tex]h_1=14\\\\h_2=\dfrac{14}{3} \\\\h_3=\dfrac{14}{9} \\\\h_4=\dfrac{14}{27} \\\\[/tex]

[tex]d=14+2*\dfrac{14}{3} +2*\dfrac{14}{9} +2*\dfrac{14}{27} \\=14*(1+\dfrac{1}{3}+\dfrac{2}{9} +\dfrac{2}{27} )\\=14*\dfrac{53}{27} \\=\dfrac{742}{27} \\[/tex]

The total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

What is the total distance?

Distance is a numerical measurement of how far apart objects or points are. It is the actual length of the path travelled from one point to another.

Here given that,

A ball is dropped from a height of [tex]14[/tex] ft. The elasticity of the ball is such that it always bounces up one-third the distance it has fallen.

So, after striking with the ground it covers the distance [tex]14[/tex] ft. so it rebounds to the height is [tex]\frac{1}{3}(14)[/tex].

Then again it hits the ground and covers the distance  [tex]\frac{1}{3}(14)[/tex] and again after rebounding it goes to the height is

[tex]\frac{1(1)}{3(3)}.(14)=\frac{(1)^2}{(3)^2}(14)[/tex]

Then it falls the same distance and goes back to the height

[tex]\frac{1}{3}[/tex] ×[tex](\frac{(1)^2}{(3)^2})[/tex] ×[tex]14[/tex] = [tex]\frac{(1)^3}{(3)^3}(14)[/tex]

So, the total distance travelled is

[tex]14+2[\frac{1}{3}(14)+(\frac{1}{3})^2(14)+(\frac{1}{3})^3(14)+...][/tex]

We take the sum is twice because it goes back to the particular height and falls to the same distance.

[tex]S=14+2(\frac{\frac{1}{3}(14)}{1-\frac{1}{3}})\\\\\\S=\frac{a}{1-r}\\\\\\S=14+2(\frac{\frac{14}{3}}{\frac{2}{3}})\\\\S=14+2(\frac{14}{2})\\\\S=14+2(7)\\\\S=14+14\\\\S=28ft[/tex]

Hence, the total distance the ball has traveled at the instant it hits the ground the fourth time [tex]28ft.[/tex]

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Diện tích xung quanh của hình chóp tứ giác đều có cạnh bằng 6cm và độ dài trung đoạn bằng 10cm là:
A. 120 cm2 B. 240 cm2 C. 180 cm2 D. 60 cm2

Answers

Answer:

B. 240 cm2

Step-by-step explanation:

Chu vi đáy: 10x=40

Diện tích xung quanh:  Sxq=1/2 x40x12=240

power sharing helps the ruling party to retain power for a long time. tick or wrong

Answers

Power tick wrong long time

d is none of the above , and yes

Answers

Answer:

[tex] = 2 {}^{2} - 3(2) = - 2 \\ 3 {}^{2} - 3(3) = 0 \\ 4 {}^{2} - 3(4) = 4 \\ 5 {}^{2} - 3(5) = 10[/tex]

Select the correct answer.
What is the best way to describe a theme of this poem?
A.
The main purpose of having New Year's resolutions is to make people feel bad.
B.
The failures of the past should inspire people to accomplish more in the future.
OC.
By the end of the year, it is too late to make any changes to a person's life.
D.
People would accomplish their New Year's resolutions if they wrote them down.

Answers

B.The failures of the past should inspire people to accomplish more in the future.

Help please guys thanks

Answers

Answer:

D

Step-by-step explanation:

sqrt_{4}(81)^5=(81^(5))^(1/4)=81^(5/4)

Answer:

D

Step-by-step explanation:

if it was properly typed, it would have been All of the above but the most correct option is D.

The management of a large airline wants to estimate the average time after takeoff taken before the crew begins serving snacks and beverages on their flights. Assuming that management has easy access to all of the information that would be required to select flights by each proposed method, which of the following would be reasonable methods of stratified sampling?

a. For each day of the week, randomly select 5% of all flights that depart on that day of the week.
b. Divide all flights into the following 4 groups on the basis of scheduled departure time: before 9:00 am 9:00 am to 1:00 pm. 1:00 pm to 5:00 pm, and after 5:00 pm. Randomly select 5% of the flights in each group.
c. For each crew member the airline employs, randomly select 5 flights that the crew member works.
d. Divide the airports from which the airline's fights depart into 4 regions: Northeast, Northwest Southwest and Southeast. Randomly select 5% of all flights departing from airports in each region

Answers

Answer:

The answer is "Option a, Option b, and Option d".

Step-by-step explanation:

In the given question it is used to stratifying the sampling if the population of this scenario it flights takes off when it is divided via some strata.

In option a, In this case, it stratified the sampling, as the population of planes taking off has been divided into the days of the week. In option b, It also used as the case of stratified sampling. In options c, it is systematic sampling, that's why it is wrong. In option d,  It is an example of stratifying the sampling.

Answer:

For each day of the week, randomly select 5% of all flights that depart on that day of the week.

Divide all flights into the following 4 groups on the basis of scheduled departure time: before 9:00 am, 9:00 am to 1:00 pm, 1:00 pm to 5:00 pm, and after 5:00 pm. Randomly select 5% of the flights in each group.

Divide the airports from which the airline's flights depart into 4 regions: Northeast, Northwest, Southwest, and Southeast. Randomly select 5% of all flights departing from airports in each region.

Step-by-step explanation:

ll sampling methods that divide the flights into a small number of mutually exclusive categories are appropriate. These methods include all flights on the basis of a characteristic that might be associated with the variable being investigated and randomly selects a proportionate number of flights from each group.

How many ways can a president, vice president, secretary, and treasurer be chosen from a club with 8 member

Answers

Answer:

504

Step-by-step explanation:

What type of line is PQ?
A. altitude
B. angle bisector
C. side bisector
D. median

Answers

the answer for your question is altitude

The line PQ of the triangle is an altitude. The correct option is A.

What is the altitude of the triangle?

A line segment passing through a triangle's vertex and running perpendicular to the line containing the base is the triangle's height in geometry.

The extended base of the altitude is the name given to this line that contains the opposing side. The foot of the altitude is the point at where, the extended base and the height converge.

In the given triangle the line segment PQ is passing through a triangle's vertex and running perpendicular to the line containing the base is the triangle's height in geometry.

Therefore, the line PQ of the triangle is an altitude. The correct option is A.

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Which number would be rounded UP to the nearest ten but DOWN to the nearest hundred?
A. 232
B. 238
C. 262
D. 268

Answers

Answer:

B

Step-by-step explanation:

b! i hope this helps

8/11 - Applying the Distributive Property
1. What is the simplest expression equivalent to 5(2x - 13) ?

Answers

10x-65
Distributive property just means multiply the 5 by each term inside the parentheses.

A ball is thrown in air and it's height, h(t) in feet, at any time, t in seconds, is represented by the equation h(t)=−4t2+16t. When is the ball higher than 12 feet off the ground?
A. 3 B. 1 C. 1 D. 4

Answers

Hence the time that the ball will be height than 12 feet off the ground is 4secs

Given the expression for calculating the height in  feet as;

h(t) = -4t²+16t

If the ball is higher than 12feet, h(t) > 12

Substituting h = 12 into the expression

-4t²+16t > 12

-4t²+16t - 12 > 0

4t²- 16t + 12 > 0

t²- 4t + 3 > 0

Factorize

(t²- 3t)-(t + 3) > 0

t(t-3)-1(t-3) > 0

(t-1)(t-3)>0

t > 1 and 3secs

Hence the time that the ball will be height than 12 feet off the ground is 4secs

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guys help me I really need your help​

Answers

Answer:

a x^2/2 is a polynomial because the power of x is 2 which is a positive whole number but 2/x^2 is not a polynomial because the power of x is -2 which is negative whole number.

b.in

[tex] \sqrt{2 x} [/tex]

the power if x will be

[tex]x {}^{ \frac{1}{2} } [/tex]

which is not a whole number so it is not a polynomial.

but in

[tex] \sqrt{2} x[/tex]

the power if x is a positive whole number.so it is a polynomial.

c.the greatest power of variable of the term is called degree of polynomial

A car travels at a constant speed towards a town. If it increases its speed by 15 km/h, the time required is in a ratio of 6 : 5. If it reduces its speed by 15 km/h, it needs another 105 minutes to arrive at the destination. Find the distance travelled by the car.

Answers

9514 1404 393

Answer:

  525 km

Step-by-step explanation:

Let d represent the distance to the town. Let s represent the nominal speed of the car. The relation between time, speed, and distance is d = st.

  t1 = d/s

  t2 = d/(s+15)

  t1 : t2 = 6 : 5 . . . increasing the speed reduces the time

Substituting for t1 and t2, we have ...

  (d/s)/(d/(s+15)) = 6/5

  (s +15)/s = 6/5

  1 +15/s = 1 +1/5

  s = 5·15 = 75 . . . . nominal speed in km/h

__

Decreasing the speed increases the time.

  d/75 +(105/60) = d/(75-15)

  d(60/75) +105 = d . . . . . . multiply by 60

  105 = d/5 . . . . . . . . . . . subtract 4/5d

  525 = d . . . . . . . . . . multiply by 5

The distance traveled by the car is 525 km.  

Here is a number sequence. The rule for finding the next term is to add
a, where a is an integer. ! ! 8 ........! ! ........! ! 29 Work out the two
missing terms.

Answers

Answer:

8,15,22,29

Step-by-step explanation:

the interger a is 7,so to find the next term you have to add 7 plus the 8,

8+7=15

15+7=22

22+7=29

8,15,22,29

I hope this helps

write your answer as an integer or as a decimal rounded to the nearest tenth​

Answers

Answer:

123456-6-&55674

Step-by-step explanation:

rdcfvvzxv.

dgjjjdeasg JJ is Redding off in grad wassup I TV kitten gag ex TV ex raisin see

recall see

HELP HELP! I NEED URGENT HELP WITH THIS equashin.

Answers

Answer:

V = 1071.79 yd^3

Step-by-step explanation:

The volume of a cone is

V = 1/3 pi r^2 h  where r is the radius and h is the height

We are given a diameter of 16 so the radius is 1/2 of the diameter or 8

The height is 16

V = 1/3 ( 3.14) (8)^2 ( 16)

V = 1071.78666 yd^3

Rounding to the nearest hundredth

V = 1071.79 yd^3

[tex] \large\begin{gathered} {\underline{\boxed{ \rm {\red{Volume \: of \: Cone \: = \: \pi \: {r}^{2} \: \frac{h}{3} }}}}}\end{gathered}[/tex]

[tex] \bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{Diameter}{2} \\ [/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{16 \: yd}{2} \\ [/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{ \cancel{16 \: yd} \: \: ^{8} }{ \cancel{2}} \\ [/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: \large{\bf{{{\color{navy}{r \: = \: 8 \: yd}}}}}[/tex]

[tex]\bf{\red{ \longrightarrow}} \tt \: \: \large{\bf{{{\color{navy}{h \: = \: 16 \: yd}}}}}[/tex]

[tex] \bf \large \longrightarrow \: \: 3.14 \: \times \: {8}^{2} \: \times \: \frac{16}{3} \\ [/tex]

[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{16}{3} \\ [/tex]

[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{ \cancel{16} \: \: ^{5.33} }{ \cancel{3}} \\ [/tex]

[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: 5.33[/tex]

[tex]\bf \large \longrightarrow \: \:200.96 \: \times \: 5.3[/tex]

[tex]\bf \large \longrightarrow \: \:1071.79[/tex]

Option (A) is the correct answer

2/8 of a rope is 28 meters.What is the length of the rope? A.32 B.42 C.4 D.21​

Answers

let length be x

ATQ

[tex]\\ \sf\longmapsto \dfrac{2}{8}\times x=28[/tex]

[tex]\\ \sf\longmapsto \dfrac{2x}{8}=28[/tex]

[tex]\\ \sf\longmapsto \dfrac{x}{4}=28[/tex]

[tex]\\ \sf\longmapsto x=4(28)[/tex]

[tex]\\ \sf\longmapsto x=112[/tex]

Step-by-step explanation:

there is something wrong with your problem description.

the offered answer options do not fit to the solution as it is described.

2/8th of a rope is 28 meters long. how long is the whole rope ?

as the other answer said : 2/8 = 1/4

and 1/4th of the rope x = 28 m

1/4 × x = 28

x (the length of the whole rope) = 4×28 = 112 meters

but - maybe the original problem said that 7/8th (and not 2/8th) of a rope is 28 m.

7/8 × x = 28

1/8 × x = 4

x = 32 m

then A (32) would be the right answer !

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