Help on A and B please

14. Find the distance between (7,217pi/180 ) and (5,-23pi/36 ) on the polar plane.

irina [24]

Answer: the distance is 3.49 units

Step-by-step explanation:

There are some ways to find the exact distance, i will calculate the distance in rectangular coordinates.

When we have a point (R, θ) in polar coordinates, we can transform it into rectangular coordinates as:

x = R*cos(θ)

y = R*sin(θ)

Then we have:

(7,217pi/180 )

R = 7

θ = (217/180)*pi

x = 7*cos( (217/180)*pi) = -5.59

y = 7*sin( (217/180)*pi) = -4.21

So this point is (-5.59, -4.21) in rectangular coordinates.

And the other point is (5,-23pi/36 )

R = 5

θ = -(23/36)*pi

x = 5*cos( -(23/36)*pi ) = -2.11

y = 5*sin( -(23/36)*pi ) = -4.53

So this point is (-2.11, - 4.53)

Then the point distance between those points is:

D = I (-2.11, -4.53) - (-5.59, -4.21) I

D = I (-2.11 + 5.59, -4.53 + 4.21) I

D = I (3.48, -0.32) I = √( (3.48)^2 + (-0.32)^2) = 3.49

4 0

3 years ago

Point O is between points Mand N on line segment MN. O is the midpoint of line segment MN if

s2008m [1.1K]

Diagram:
M------------O-------------N
................. ^ midpoint
Obviously if O is the *mid*point, then it is exactly halfway between M and N. That also means it has bisected the segment into two equal parts.
On one side you have MO (or OM, if you like)
On the other side you have ON.
In order for O to be the midpoint:
ON = OM

8 0

3 years ago

Plzzzzz someone help me

solong [7]

The answer is The equation 3.5|6x – 2| = 3.5 has one solution

4 0

2 years ago

Read 2 more answers

A candy box is made from a piece of cardboard that measures 23 by 13 inches. Squares of equal size will be cut out of each corne

wolverine [178]

Answer:

2.67 inches.

Step-by-step explanation:

Assuming that we represent the size of the squares with the letter y, such that after the squares are being cut from each corner, the rectangular length of the box that is formed can now be ( 23 - 2y), the width to be (13 - 2y) and the height be (x).

The formula for a rectangular box = L × B × W

= (23 -2y)(13-2y) (y)

= (299 - 46y - 26y + 4y²)y

= 299y - 72y² + 4y³

Now for the maximum volume:

dV/dy = 0

This implies that:

299y - 72y² + 4y³ = 299 - 144y + 12y² = 0

By using the quadratic formula; we have :

$= \dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$