type the equation that shows the relationship between the variables in this chart: x: 1,2,3,4,5 and y: 7,14,21,28,35

Answers

Answer 1

Answer: The equation that relates the x and y values in the chart above is: y = 7x. It can be observed that each value of x when multiplied with a factor of 7 gives the corresponding y value. For example, for x equals 1, y is equal to 7 times 1 which gives a value of 7. For x equals 5, the y value is equal to 7 times 5 which gives the number 35.

Answer 2

Answer:

Final answer:

The chart displays a linear relationship between x and y, represented by the equation y = 7x. Here, x is the independent variable, with y changing dependent on x's value.

Explanation:

The relationship between the variables x and y in the chart can be represented by the equation y = 7x. This is a type of linear equation. In this formula, x is the independent variable, and y is the dependent variable. The value of y depends on the value of x. You can check the validity of the equation by substituting the value of x from the chart into the equation to obtain the corresponding y-values. For example, when x = 1, y = 7*1 = 7, which matches the corresponding y value in the chart. This process can be repeated for all x values.

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Ms. White has 15 students in her first grade class. Troy is the line leader for the week, and Mackenzie is last because she was the line leader last week. In how many different ways can Ms. Whites class line up for lunch?A.) 13!
B.) 15!
C.) 15!÷2!

Answers

The first place and 15th place are already decided, so we have to find the number of different ways that the other 13 students can line up, in the places from 2 to 14.

2nd place can be any one of 13 people. For each of those . . .

3rd place can be any one of 12 people. For each of those . . .

4th place can be any one of 11 people. For each of those

13th place can be any one of 2 people. For each of those

14th place has to be the one student who is left.

Total number of ways that 13 students can line up in places 2 through 14 is

(13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1)

That number is called "thirteen factorial". The number is 6,227,020,800 .

When you write it in math, you write it like this:13

Answer:

13!

Step-by-step explanation:

My teacher gave us an assignment it says D=S/T 35days = _ weeks

Answers

Answer:

To convert 35 days into weeks using the formula D = S / T, where:

- D is the time duration in days (35 days).

- S is the time duration in weeks (what you want to find).

- T is the conversion factor from days to weeks (7 days per week).

You have D = 35 days, and you want to find the equivalent duration in weeks (S).

So, rearrange the formula to solve for S:

S = D / T

Now, plug in the values:

S = 35 days / 7 days/week

S = 5 weeks

So, 35 days is equivalent to 5 weeks.

The circumference of the question has established a voluminous magnitude of our resplendent celestial orb, known as the Sun, is empirically approximated to be a staggering 1.40927256905986 x 10^18 cubic kilometers when expressed in the exalted parlance of scientific notation, elegantly rendered with a precision of precisely six significant digits. This numerically magnificent representation, denoted in its abbreviated splendor as 1.409 x 10^18 km³, befits the celestial colossus that reigns supreme within our solar dominion.

Find the measure of each interior angle of a regular decagon.The measure of each interior angle is
°.

Answers

Answer:

The measure of each interior angle is 144°

Step-by-step explanation:

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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Choose the less precise measurement and explain why it is less precise. 2.8 mm or 3.79 mm

Answers

2.8 MM because it is probably an estimate because nothing is perfect if you zoom in close enough!

Jane is making thanksgiving cookies while her daughter mary practices for a maths test. to test her, jane makes her cookies in5 different shapes and groups them according to a logic puzzle. which shape completes the fourth group: triangle, pentagon, square; square, hexagon, hexagon, square; pentagon, hexagon, hexagon, hexagon, square, triangle; hexagon, octagon, octagon, octagon, octagon, ________.

Answers

square square square square

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