MATHEMATICS HIGH SCHOOL

10 kilometers is about
6 miles
B
6 feet
10 feet
60 feet

Answers

Answer 1
Answer: 6 miles because 1km is .6214 miles, 10x.6214 ≈ 6
Answer 2
Answer: Your answer is 6miles !

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Help me please just click on the picture

Plz help !!!!!!!!!! If correct I will give brainiest

Answers

Answer:

Step-by-step explanation:

the y intersept is 2

so it should be 1 x 2

A and b have the same direction, and they both have a magnitude of 6. What must be true about a and b ? a. They are equal and parallel.
b. They are opposites, but not parallel.
c. They are opposites and parallel.
d. They are equal, but not parallel.

Answers

Answer: Option a

They are equal and parallel

Step-by-step explanation:

If a and b have the same direction then necessarily a and b are parallel.

We know that they also have the same magnitude.

Two vectors are equal if they have the same magnitude and direction.

Then we can say that a and b are equal and we can also say that they are parallel.

Therefore the correct option is the option a

Answer:

A

Step-by-step explanation:

Edge2021

Suppose you average 52 mi/h traveling on the highway. If you drive for 5 hours, how far will you travel? A. 260 miles B. 250 miles C. 350 miles D. 240 miles

Answers

if you average 52 mi/h for 5 hours you would travel about 260 miles
A. 260 miles. 52 per hour, so 52 times 5.

Answer the following questions about the problem above. Write in complete sentences to get full credit. 1. What is the slope for section "d" of Mrs. Washington's commute.
2. What does it mean that the slope is negative in context of the problem?
3. Why are the slopes different over different intervals?
4. How long does it take Mrs. Washington to get home? How did you know this?

Answers

Note: Since you missed to add the image, so after a little research, I was able to find the image related to this question, which hopefully serves the purpose to clear your concept. The image is attached below.

Step-by-step explanation:

Part 1) What is the slope for section "d" of Mrs. Washington's commute

Calculating the slope for section "d" of Mrs. Washington's commute:

Slope of the section d  =(y_(2) - y_(1))/(x_(2) - x_(1))

(6 - 0)/(32 - 20)=(1)/(2)=0.5

Part 2) What does it mean that the slope is negative in context of the problem?

The slope is being negative as it is moving downward from the left. In negative slop, the value of x increase, while the vale of y decreases. In positive slop, the value of both x and y increases.

Part 3) Why are the slopes different over different intervals?

Slopes are different at different intervals as the distance covered in time taken for each part of the given journey are different, also the speed is not the same for each interval.

As the speed can be calculated as

Speed\:=\:(distance)/(time)\:

Distance = y_(2) -y_(1) = 20 - 15 = 5

Time = x_(2) -x_(1)= 8 - 0 = 8

Speed = (5)/(8)=0.625

Therefore, the speed in the part 1 is 0.625ms^(-1)

Part 4) How long does it take Mrs. Washington to get home? How did you know this?

It is clear from the graph that does take 32 minutes to her to get home.  

We can observe it from the graph that it represents the total time taken across the entire journey.

Keywords: slope, speed, time

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A field goal kicker lines up to kick a 44 yard (40m) field goal. He kicks it with an initial velocity of 22m/s at an angle of 55∘. The field goal posts are 3 meters high.Does he make the field goal?What is the ball's velocity and direction of motion just as it reaches the field goal post

Answers

Answer:

The ball makes the field goal.

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of motion is -45.999º or 314.001º.

Step-by-step explanation:

According to the statement of the problem, we notice that ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical uniform accelerated motion, whose equations of motion are described below:

x = x_(o)+v_(o)\cdot t\cdot \cos \theta(Eq. 1)

y = y_(o) + v_(o)\cdot t \cdot \sin \theta +(1)/(2)\cdot g\cdot t^(2)(Eq. 2)

Where:

x_(o), y_(o) - Coordinates of the initial position of the ball, measured in meters.

x, y - Coordinates of the final position of the ball, measured in meters.

\theta - Angle of elevation, measured in sexagesimal degrees.

v_(o) - Initial speed of the ball, measured in meters per square second.

t - Time, measured in seconds.

If we know that x_(o) = 0\,m, y_(o) = 0\,m, v_(o) = 22\,(m)/(s), \theta = 55^(\circ), g = -9.807\,(m)/(s) and x = 40\,m, the following system of equations is constructed:

40 = 12.618\cdot t(Eq. 1b)

y = 18.021\cdot t -4.904\cdot t^(2)(Eq. 2b)

From (Eq. 1b):

t = 3.170\,s

And from (Eq. 2b):

y = 7.847\,m

Therefore, the ball makes the field goal.

In addition, we can calculate the components of the velocity of the ball when it reaches the field goal post by means of these kinematic equations:

v_(x) = v_(o)\cdot \cos \theta(Eq. 3)

v_(y) = v_(o)\cdot \cos \theta + g\cdot t(Eq. 4)

Where:

v_(x) - Final horizontal velocity, measured in meters per second.

v_(y) - Final vertical velocity, measured in meters per second.

If we know that v_(o) = 22\,(m)/(s), \theta = 55^(\circ), g = -9.807\,(m)/(s) and t = 3.170\,s, then the values of the velocity components are:

v_(x) = \left(22\,(m)/(s) \right)\cdot \cos 55^(\circ)

v_(x) = 12.619\,(m)/(s)

v_(y) = \left(22\,(m)/(s) \right)\cdot \sin 55^(\circ) +\left(-9.807\,(m)/(s^(2)) \right)\cdot (3.170\,s)

v_(y) = -13.067\,(m)/(s)

The magnitude of the final velocity of the ball is determined by Pythagorean Theorem:

v =\sqrt{v_(x)^(2)+v_(y)^(2)}(Eq. 5)

Where v is the magnitude of the final velocity of the ball.

If we know that v_(x) = 12.619\,(m)/(s) and v_(y) = -13.067\,(m)/(s), then:

v = \sqrt{\left(12.619\,(m)/(s) \right)^(2)+\left(-13.067\,(m)/(s)\right)^(2) }

v \approx 18.166\,(m)/(s)

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of the final velocity is given by this trigonometrical relation:

\theta = \tan^(-1)\left((v_(y))/(v_(x)) \right)(Eq. 6)

Where \theta is the angle of the final velocity, measured in sexagesimal degrees.

If we know that v_(x) = 12.619\,(m)/(s) and v_(y) = -13.067\,(m)/(s), the direction of the ball is:

\theta = \tan^(-1)\left((-13.067\,(m)/(s) )/(12.619\,(m)/(s) ) \right)

\theta = -45.999^(\circ) = 314.001^(\circ)

The direction of motion is -45.999º or 314.001º.

The ball makes the field goal.

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of motion is -45.999º or 314.001º.

According to the statement of the problem, we notice that ball experiments a parabolic motion, which is a combination of horizontal motion at constant velocity and vertical uniform accelerated motion, whose equations of motion are described below:

X=Xo+Vo*t*cosФ (Eq. 1)

Y=Yo+Vo*t*sinФ +(1/2)*g*t²(Eq. 2)

Where:

Xo,Yo  - Coordinates of the initial position of the ball, measured in meters.

X,Y  - Coordinates of the final position of the ball, measured in meters.

Ф- Angle of elevation, measured in sexagesimal degrees.

Vo - Initial speed of the ball, measured in meters per square second.

t - Time, measured in seconds.

If we know that Xo = 0m, Yo = 0m, Vo = 22m/s, Ф = 55°,g = -9.807m/s  and X = 40m, the following system of equations is constructed:

40 = 12.618*t (Eq. 1b)

Y = 18.021*t-4.904*t² (Eq. 2b)

From (Eq. 1b):

t = 3.170s

And from (Eq. 2b):

Y = 7.847m

Therefore, the ball makes the field goal.

In addition, we can calculate the components of the velocity of the ball when it reaches the field goal post by means of these kinematic equations:

Vx = Vo*cosФ (Eq. 3)

Vy = Vo*cosФ+g*t (Eq. 4)

Where:

Vx - Final horizontal velocity, measured in meters per second.

Vy- Final vertical velocity, measured in meters per second.

If we know that Vo = 22m/s, Ф= 55°, g = -9.807m/s  and t = 3.170s, then the values of the velocity components are:

Vx = (22m/s)*cos55°

Vx = 12.619m/s

Vy = (22m/s)*sin55°+(-9.807m/s²)*3.170s

Vy = -13.067m/s

The magnitude of the final velocity of the ball is determined by Pythagorean Theorem:

V = √(Vx²+Vy²) (Eq. 5)

Where  is the magnitude of the final velocity of the ball.

If we know that  Vx = 12.619m/s and Vy = -13.067m/s, then:

V = √((12.619m/s)²+(-13.067m/s)²)

V ≈ 18.166m/s

The magnitude of the velocity of the ball is approximately 18.166 meters per second.

The direction of the final velocity is given by this trigonometrical relation: Ф = tan^(-1)(Vy/Vx)(Eq. 6)

Where Ф is the angle of the final velocity, measured in sexagesimal degrees.

If we know that  Vx = 12.619m/s and Vy = -13.067m/s, the direction of the ball is:

Ф = tan^(-1)((-13.067m/s)/(12.619m/s))

Ф = -45.999° = 314.001°

The direction of motion is -45.999º or 314.001º.

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Which of the following fractions will result in a repeating decimal? A. 1/3 B. 4/5 C. 3/8 D. 1/2

Answers

Answer:

A .   1/3

Step-by-step explanation:

1/3 = 0.333333333333333 (and the 3 continues on and on and on and on)

4/5 = 0.8

3/8 = 0.375

1/2 = 0.5
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