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Let's find out :

6x+30+4x = 10(x+3)

10x+30 = 10x+30                 ANY value of x will make this true...infinite answers !

how many solutions are in this equation.

6x+30+4x=10(x+3)

A. 1

B. Infinitely many

C. 0

A guess would be only one as it is an equation of x¹

^{Try it out:}

^{8x+47 = 8x + 40}

^{47=40                                    This is NOT true...so there are actually   ZERO solutions !}    

In the first scenario, h will be a positive number

in the second drawing h = 0

in the third scenario h will be some negative number

.02 d^2 - 0.6 d +1     will have a nadir (a low point) at  -b/2a   =  -  -0.6/(2*.02) = 15 m = d

SUstitute this value in for d    .02(15^2) - 0.6(15) + 1 = h  = = -3.5    so this applies to the third picture.

DO the same thing for this equation   a (low point) = - b/ 2a = 15  (again)

    now substitute this value into the equation and find   h   to see what scenario this second equation fits....... you can do it!  

Draw a square of some side

Draw a rhombus with the same side length(s)  as the square

The figures will be in proportion but they wll not be similar  !!!

= 6x^2 + 8x + 2   

Just imagine two  polygons made out of pipe cleaners......and all of the sides are proprtional (or even the same length)....then you knda squoosh one to the side.....now it has equal length sides, but not angles and it doesn't look like the other....can you draw something like that?

3(4x+6)=9x+12

12x + 18   =  9x + 12       subtract 9x, 18 from both sides

3x = - 6       divide both sides by 3

x = -6 / 3   =   -2

Excellent, EW!!!!

A 2 : 3 ratio means that this point is (2/5) of the way from A to B

We can find the point thusly :

 [ 2 + (6 - 2)(2/5) , 7 + (2 - 7)(2/5) ] =

[ 2 + (4)(2/5)  , 7 + (-5)(2/5) ] =

[ 2 + 8/5  , 7 - 2   ]    =

( 10/5 + 8/5 , 5 ) =

(18/5, 5)

The locations of the friends will form vertexes of a triangle

If  we construct perpendicular bisectors of each side of this triangle.....they willl neet at the circumcenter.....

This will be the center of a circle that will pass  through each vertex....so.....the radial distance from this point to each of his friends will be equal

[BTW...this point could lie outside the triangle   !!! ]

Here's a good resource to show this : http://www.mathopenref.com/trianglecircumcenter.html

14n + 6p - 8n = 18p

First group the ike terms

6p + 14n - 8n = 18p

6p + 6n = 18p

Subtract 6p on both sides

6p + 6n - 6p = 18p - 6p

6n = 12p

Then divide 6 on both sides

6n/6 = 12p/6

n = 2p

C is the answer

Hope this helps ;P

You need to draw a line through the apex point of the triangle that is parallel to the triangle's base

This shows how to do this :

http://www.mathopenref.com/printparallel.html

We can then use the fact that alternate interior angles are equal to complete the proof

If angle OAL =  70°   GO would be parallel to LA

To see this.....extend LA in the direction of A to   E

Then angle OAL + angle  OAE  sum to 180°

But angle OAE =  180 - angle OAL   =  180 - 70 =  110

Then angles GOA and OAE are equal

And if this is so....then transversal OA makes alternate interior angles GOA and OAE equal....so....

A transversal cutting lines (segments) so that alternate interior angles are equal means that the lines (segments) are parallel

So...GO is parallel to LA.....and a trapezoid must have one pair of sides parallel

I really don't understand why you insist on solving all these problems to completion.

Do you want the students to cheat? Because that's what you're doing.  Helping students to cheat.

(961, 1147, 1271, 1333, 1369, 1457, 1517, 1591, 1643, 1681, 1739, 1763, 1829, 1849, 1891, 1927, 1961, 2021, 2077, 2173, 2183, 2201, 2209, 2257, 2263, 2279, 2419, 2449, 2479, 2491, 2501, 2537, 2573, 2623, 2627, 2701, 2747, 2759, 2773, 2809, 2867, 2881, 2911, 2923, 2993, 3007, 3053, 3071, 3127, 3139, 3149, 3233, 3239, 3293, 3337, 3397, 3403, 3431, 3481, 3551, 3569, 3589, 3599, 3649, 3713, 3721, 3763, 3827, 3869, 3901, 3953, 3977, 4087, 4171, 4183, 4187, 4189, 4307, 4331, 4399, 4453, 4489, 4559, 4661, 4717, 4757, 4819, 4891, 4897, 5041, 5063, 5141, 5183, 5251, 5293, 5329, 5429, 5561, 5609, 5723, 5767, 5893, 5917, 5963, 6059, 6241, 6319, 6497, 6499, 6557, 6887, 6889, 7031, 7081, 7387, 7663, 7921, 8051, 8633, 9409) = 120 such numbers.
ALL these numbers have Prime Factors that fall between 31 and 97. There are 15 Prime Numbers that form the basis of the 120 numbers above as follows:
31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97. And the above 120 numbers range between 31^2 and 97^2, and are calculated as follows:
15(nCr)2 =105 + 15 squares(31^2, 37^2, 41^2.....and so on = 120 numbers.

A slight adjustment in the answer above:

sumfor(n, 1, 99, (1/((n+1)^3 - (n+1)) = 0.249950495 

Segment PT || segment QS,                                  Given

segment PT ≅ segment QS,

∠T ≅ ∠S

Angle TPQ = Angle SQR                      PR is a transversal  cutting parallel segments SQ and TP

                                                              So....it makes corresponding angles TPQ and SQR equal   

ΔPQT   ≅    ΔQRS                                ASA congruency

Draw BD                         A segment connects any two points

AB = BC                        Properties of a kite - two pairs of equal adjacent sides

AD = CD 

BD = BD                        Reflexive Property

ΔABD  congruent to         SSS

ΔCBD

∠A ≅ ∠C                         Congruent parts of congruent triangles are themselves congruent

sumfor(n, 1, 100, (1/((n+1)^3 - (n+1)) = 0.2499514657

a)   Profit  at 4.5 dollars =   -15(4.5)^2 + 195(4.5) -450 = $123.75

b) The intersection  with the horizontal axis is where

-15p^2 + 195p - 450 = 0      multiply through by -1

15p^2 - 195p + 450 = 0     divide through by 15

p^2 - 13p  + 30   =  0       factor

(p - 10) (p - 3) = 0

Set each factor to 0  and solve for p we get that  p=  3   and p = 10

c)  These points mean that she will earn a profit as long as the price is between $3 and $10

d) As Rom found....p = 13/2 = 6.5 dollars earns the max profit

And this max profit is   -15(6.5)^2 + 195(6.5) - 450 = $183. 75

4x^2 + 20x + 24