Convert 123cg to kg?

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Answer 1

Answer: Just add two zeros to the front to make it kg .00123

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For the coordinates of P and M (the midpoint) are given. Find the coordinates for Q.P(3, 11) and M (0, 0)

Answers

Given an endpoint of a segment and a midpoint, the other endpoint can be obtained by manipulation of the midpoint formula. The said formula is shown below:

Let: (a,b) = coordinates of point 1 ; (c,d) = coordinates of point 2; (e,f) = coordinates of the midpoint

Midpoint = ( (a+c)/2 , (b + d)/2 )

From the formula: (a+c)/2 = e ; (b + d)/2 = f

Since we are already given an endpoint and the midpoint, we can solve for the other endpoint using the obtained equations. This is shown below:

(a+c)/2 = e
(3 + c)/2 = 0
c = -3

(b + d)/2 = f
(11 + d)/2 = 0
d = -11

Therefore, the coordinates of the other point is Q(-3,-11)

Please help me... I don't understand why I got it wrong(

Answers

You did the first step correctly ... even admirably !
You zeroed in on finding 'x' with laser-like precision, and
you knocked it out of the park.
Then, from there ... you fell off the deep end.

There's no reason that (3x + 18) has to be equal to 2(y+6).
In fact, they add up to 180°, so you could have used that
and solved for 'y'.

BUT ... (5x + 2) and 2(y+6) are "vertical angles", and THEY
are equal.

So 2 (y+6) = 5x + 2

2 (y+6) = 102

(y+6) = 51

y = 45 .

You got the answer y=78 by substituting x=20 into the top angle, 3x+18.
But the angle 3x+18 is not equal to the angle with the y variable.
They are supplementary angles (equal 180) but are not necessarily equal.
The angle that equals 2(y+6) is the angle opposite to it, 5x+2 (Vertical angle theorem).
So you should have substituted 20 into 5x+2

Solve of system of equation.
9x + 2y = -38
9x+10y=26.

Answers

Final answer:

The system of equations can be solved using the method of elimination. The solution is x = -6 and y = 8.

Explanation:

To solve this system of equations, we can use the method of elimination. First, we'll multiply the first equation by 10 and the second equation by 2 to make the coefficients of x the same. This gives us:

90x + 20y = -380
18x + 20y = 52

Now, we can subtract the second equation from the first to eliminate y:

72x = -432

Dividing both sides by 72, we find x = -6.

Substituting this value of x into one of the original equations, we can solve for y:

9(-6) + 2y = -38

Simplifying, we get -54 + 2y = -38

Adding 54 to both sides, we have 2y = 16

Dividing both sides by 2, we find y = 8.

Learn more about Solving systems of equations here:

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Answer:

(-6 , 8 )

Hope this helps!

A boat costs $19,200 and decreases in value by 12% per year. How much will the boat be worth after 5 years?

Answers

Find 10% of $19,200 = 1,920
Find 2% of $1,920 = 384
12% = $2,304
19,200 - 2,304 = 16,896 (year 1)
Repeat this until you have your 5th year :)

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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Find two whole numbers whose sum is 60 and whose difference is 12

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$\left \{ {\big{x+y=60} \atop \big{x-y=12}} \right. \n--------\nx+x=60+12\n2x=72\ /:2\nx=36\n\nx+y=60\ \ \ \Rightarrow\ \ \ y=60-x\ \ \ \Rightarrow\ \ \ y=60-36=24\n\nAns.\ the\ numbers:\ 36\ \ and\ \ 24$

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