A chord can contain at most how many diameters

Answers

Answer 1

Answer: A cord links any two points on a circle.

The greatest distance between two points on a circle is the length of the diameter.

Hence a chord can contain at most one diameter.

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A man buys 12.50kg of one of rice for £30 per kg and £67.90 per kg. If he mixes them and makes packets, each containing 3450g of rice, how many packets of rice are there and what is the cost of each packet?

Answers

Answer:

a. The number of packets the two bags/brands of rice can be repackaged to = 7 packets.

b. Cost of one packet is £174.82

Step-by-step explanation:

Please kindly see the attached files for explanation.

The graph shows two lines PLS EXPLAIN YOUR ANSWER

Answers

Answer:

None

Step-by-step explanation:

There is a solution when the two lines intersect, because the lines have the same slope (and different y intercepts) they will never intersect. Therefore there are no solutions

For 7/20, what is an equivalent fraction with a denominator of 100?

Answers

35. The denominator is 20. To make it 100, it need to times 5. So its 20x5=100. Do the same with numerator. 7x5=35

Hey there, I think this might help. If it helps please give me the brainiest answer.

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.

For more questions on magnitude.

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A triangle, ΔABC, is reflected across the x-axis to have the image ΔA'B'C' in the standard (x,y) coordinate plane; thus, A reflects to A'. The coordinates of point A are (c,d). What are the coordinates of point A'?

Answers

The coordinates of point A' is (c, -d).

What is a reflection?

There are two ways of reflection.

Along x-axis:

(x, y) – (x, -y)

Along y-axis:

(x, y) - (-x, y)

We have,

A triangle, ΔABC, is reflected across the x-axis to have the image ΔA'B'C'.

Now,

A = (c, d)

Reflection across the x-axis.

This means,

(x, y) → (x, -y)

So,

(c, d) → (c, -d)

Thus,

(c, -d) is the coordinates of point A'.

Learn more about reflections here:

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Answer:

sorry fr being 7 years late, but the answer is option a!

Step-by-step explanation:

e) Andy is 2 times younger than his sister and his father is 25 years older than him. If the total of their ages is 53 years, what is Andy’s age and his father’s age?

Answers

An equation is formed of two equal expressions. The age of Andy is 7 years, while the age of his father is 32 years.

What is an equation?

An equation is formed when two equal expressions are equated together with the help of an equal sign '='.

Let Andy's sister's age be represented by y, while the age of Andy is represented by x. Given that Andy is 2 times younger than his sister. Therefore, we can write the equation as,

Sister's age, y = 2x

Also, Andy's father is 25 years older than him. Therefore, we can write,

Father = x + 25

Since the sum of all the people's age is 53. Therefore, we can write,

Sum = y + x + Father's age

53 = 2x + x + x + 25

53 = 4x + 25

28 = 4x

x = 28/4

x = 7

Therefore, the age of the family members can be written as,

Andy's age = x = 7 years