NEED HELP ASAP PLEASE . End of quarter today

Find all of the polar coordinates for (6, 31 degrees)

Jobisdone [24]

(6, 31+2pi degrees)
(6, 31+4pi degrees)
(-6, 31-pi degrees)
(6, 31-2pi degrees)

etc. etc.

8 0

3 years ago

The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width

blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given perimeter

perimeter of rectangle = 2 ( length + width )

=> 30.8 km = 2 ( x + y )

=> $x + y = \frac{30.8}{2}$

=> y = 15.4 – x ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,

=> $length^2 + width^2 = diagonal^2$

=> $x^2 + y^2 = 11^2$

=> $x^2 + y^2 = 121$

On substituting value of y from (1) in above equation we get

=> $x^2 + (15.4-x)^2 = 121$

=> $x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x = 121$

=> $2x^2-30.8x + 237.16 -121 = 0$

=> $2x^2-30.8x + 116.16 = 0$

Solving above quadratic equation using quadratic formula

General form of quadratic equation is

$ax^2 +bx +c = 0$

And quadratic formula for getting roots of quadratic equation is

$x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}$

As equation is $2x^2-30.8x + 116.16 = 0$ , in our case

a = 2 , b = -30.8 and c = 116.16

Calculating roots of the equation we get

$x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}$

$x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}$

$x=\frac{30.8\pm\sqrt{860.64}}{4}$

$x=\frac{(30.8\pm29.33)}4$

$x=\frac{(30.8+29.33)}{4}$

$x=\frac{(30.8-29.33)}{4}$

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0

3 years ago

How do you find the equation of a triangles altitude using point slope formula?

forsale [732]

1. An altitude of a triangle is a line segment from a vertex perpendicular to the opposite side. Find the equations of the altitudes of the triangle with vertices (4, 5),(-4, 1) and (2, -5). Do this by solving a system of two of two of the altitude equations and showing that the intersection point also belongs to the third line. 
(Scroll Down for Answer!)Answer by jim_thompson5910(34047) (Show Source):You can put this solution on YOUR website!
If we plot the points and connect them, we get this triangle:

Let point
A=(xA,yA)
B=(xB,yB)
C=(xC,yC)

-------------------------------
Let's find the equation of the segment AB

Start with the general formula

Plug in the given points

Simplify and combine like terms

So the equation of the line through AB is

-------------------------------
Let's find the equation of the segment BC

Start with the general formula

Plug in the given points

Simplify and combine like terms

So the equation of the line through BC is

-------------------------------
Let's find the equation of the segment CA

Start with the general formula

Plug in the given points

Simplify and combine like terms

So the equation of the line through CA is

So we have these equations of the lines that make up the triangle

So to find the equation of the line that is perpendicular to that goes through the point C(2,-5), simply negate and invert the slope to get

Now plug the slope and the point (2,-5) into

Solve for y and simplify

So the altitude for vertex C is

Now to find the equation of the line that is perpendicular to that goes through the point A(4,5), simply negate and invert the slope to get

Now plug the slope and the point (2,-5) into

Solve for y and simplify

So the altitude for vertex A is

Now to find the equation of the line that is perpendicular to that goes through the point B(-4,1), simply negate and invert the slope to get

Now plug the slope and the point (-4,1) into

Solve for y and simplify

So the altitude for vertex B is

------------------------------------------------------------
Now let's solve the system

Plug in into the first equation

Add 2x to both sides and subtract 2 from both sides

Divide both sides by 3 to isolate x

Now plug this into

So the orthocenter is (-2/3,1/3)

So if we plug in into the third equation , we get

So the orthocenter lies on the third altitude

3 0

3 years ago

The base of a triangle is represented by (2x − 6) and the height by (x − 2). Which polynomial expression BEST represents the are

BabaBlast [244]

Area of a triangle= $\frac{1}{2} *base*height$
A= $\frac{1}{2} *(2x-6)*(x-2)$
A= $\frac{1}{2} ( 2x^{2} -4x-6x+12)$
A= $x^{2} -2x-3x+6$
A= $x^{2} -5x+6$

7 0

3 years ago

Read 2 more answers

Can someone answer this please so I can go to the back page

igomit [66]

Blake has 25 CD's.

25 times 3 = 75
75 + 7 = 82

Joel has 82 CD's.

If Joel has twice as many as Mariella, then divide 82 by 2 and you get 41.

Mariella has 41 CD's.

7 0

4 years ago