Item 12 a rectangular room is 10 feet longer than it is wide. one-foot-by-one-foot tiles cover the entire floor. let ww represent the width (in feet) of the room. write an expression in simplest form that represents the number of tiles along the outside of the room.

Answers

Answer 1

Answer:

Given:

Rectangular room is 10 feet; and

1 x 1 ft tiles

The expression for this problem would be:

If we know that the width is w, while the length is w + 10, therefore the perimeter is 2w + 2 (w + 10), so you need 4w + 20 - 4, or 4w + 16 tiles. Since you do not want to count the corners, twice.

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A teacher is grading exit tickets on the train. It takeshim 42.5 seconds to grade each exit ticket, and he will
arrive at his destination in 5 minutes. The teacher knows
he will need to save 30 seconds to pack up his materials.
What is the maximum number of exit tickets that he can
grade?

A) 8
B) 7
C) 6
D) 5

Answers

Answer:

6 papers

Step-by-step explanation:

5 minutes = 5 * 60 = 300 seconds

He needs to save 30 seconds

300 - 30 = 270 seconds

270 seconds / 42.5 seconds per paper

6.352941176 papers

Rounding down since he want to completely grade the paper

6 papers

How do I factor this polynomial? q^2-121

Answers

$q^2-121 =q^2-11^2=(q-11)(q+11)\n \n \na^2-b^2 =(a-b)(a+b)$

If 4x^2 - 100 = 0, what are the roots of the equation?

Answers

the roots to this equation i s either positive or negative five

Positive or negative 5

Write 10 12/5 as an equivalent improper fraction

Answers

10 $(12)/(5)$ = $((10 X 5 ) + 12)/(5)$
= $(62)/(5)$

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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Moira is a teacher, and she drives to school and back every day from her home. It is 32 miles from her home to work.How far does Moira drive roundtrip in 5 work days?

Answers

Answer:

Here is the answer 160 miles

Step-by-step explanation:

If you want more explanation you just do 32x5=160 or if its back and forth it would be 32x10=320

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