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The x coordinate of the vertex is    8 / (2*2)  = 2

f(-1) =  2(-1)^2 - 8(-1) + 1 = 11

f(2)  =  2(2)^2 - 8(2) + 1 =   -7

The vertex is (2,-7)

So....the range on [-1,2]  is [ -7, 11]

I used the snipping tool for this......but....Desmos has the ability to  import a graph..however....it will not import the equations  and I wanted to show those

If you have trouble, let me know and I'll help

Looks like your HW, with the last sentence and all. Why don't you try it yourself, give it some effort, and when you cannot do it, can ask for help. and put what you've tried.

For the second part, there are 18 people, and the 2 people have to be different, so its just 18 * 17 divided by 2 which is 153 ways

Umm.. What is the (18 18)? Is there suppose to be like a symbol or something? Also what is it equal too..?

Hello Cphill, do you mind telling me how to post a desmos picture? I've been answering questions and its very frustrating to only describe the picture

Find the intersection point of

2x + 5y = 10  and 3x + 4y =12

2x + 5y =10   →   6x + 15y  = 30     (1)      

3x + 4y = 12  →   -6x -8y  = - 24      (2)

Add (1) and (2)

7y = 6

y = 6/7

2x + 5(6/7)  =10

2x + 30/7 = 70 / 7

2x = 70 / 7 - 30 / 7

2x =  40  / 7

x = 40 / 14  =   20 / 7

The minimum value of  8x + 13y will occur at (20/7, 6/7) =

8 ( 20/7)  + 13 (6/7)   =

( 160  + 78)  / 7  =

238/ 7  = 

34

The probability is 6/23

Suppose that Steve chooses

Wood  Red Circle

There are  3 tiles remaining  that are both Wood and Red

Wood Red Triangle

Wood Red Square

Wood Red Pentagon

There are 2 tiles remaining that are both Wood and Circle

Wood  Yellow  Circle

Wood   Blue   Circle 

There is 1 tile remaining that  is Red and Circle

Plastic Red Circle

After Steve chooses a tile......there are 23 remaining and there will  always be 6 that share two properties in  common  with Steve's

So....the probability is 6/23

Let  B = (0,0)

C =(8,0)

Construct  a circle with a radius of  5  centered at B

The equation is  x^2 + y^2  =  25

Since AB = AC, the triangle is isosceles and A = (BC/2, y)  = (8/2, y)  = (4,y)

The x coordinate of A = 4

To find y

4^2 + y^2  =25

16 + y^2  = 25

y^2 = 9

y = 3 =   the height of ABC

So  A = (4,3)

The incenter will occur at   (4, m)

To find m

[ 5*0  + 5*0+ 8*3 ] /[5 + 5 + 8]  =   24 / 18 =  4/3

So I = (4, 4/3)

Triangles ABC and AMN are similar

[ABC]  = (1/2)(BC)(height) = (1/2) (8) (3)  =  12

Height of AMN  = 3 - 4/3 =  5/3

Scale factor of AMN to ABC  = (5/3) / 3  =  5/9

Area of AMN  = Area of ABC * (5/9)^2 =   12 (5/9)^2   =  100 / 27

A stick is broken at two points, chosen at random. If the length of the stick is 6 units, then what is the probability that all three resulting pieces are shorter than 2 units?  

Probability is zero.  If all 3 pieces are shorter than 2 units, then they don't add up to the original 6 of the stick.  

_.

Factor  2700  =  2^2 * 3^3 * 5^2  =  2*2*3*3*3*5*5

Possible Rolls                  Identifiable Sequences

22333551                         8! / ( 2! * 3! * 2!)  =   1680

43335511                         8!  /  (3! * 2! * 2!)  =  1680 

62335511                         8!  / (2! * 2! * 2!)   =  5040

66355111                         8!  /  (2! * 2! * 3!)  =  1680       

Identifiable sequences  =  3(1680) + 5040 =  10080

(0, 0)        (1,  1)

(2, 0)        (1, -1)

(-2,0)        (-1, 1)

(0,-2)        ( -1,-1)

(0, 2)

3331  =    4! / 3! =  4 integers

5311  =    4! / 2! =  12 integers

16 integers

                                             A

                            E                      D

                                  T

                B                         M                  C

Height of ABC  =  AM

Base of ABC = BC

[ ABC ]  = (1/2) (BC) (AM) =  14   → (BC) (AM)  =  28

Height of BEC   = (1/2)AM

Base of BEC =  BC

[ BEC ]  = (1/2) (BC) (1/2)AM =  (1/4)(BC) (AM)  =  (1/4) (28)  =  7

its 1

A

4                5

B           D            C

              3

Since AD bisects  angle BAC

Then  let DC=  x  and BD = 3 -x

So

(3 - x) / 4 = x / 5

5 (3 -x)  = 4x

15 - 5x = 4x

15 = 9x

x = 15/9  = 5/3

Height of triangle ADC = 4   base = (5/3)

Area =  (1/2) (5/3) 4  =  20 /  6  =  10/3  =  3  (to the nearest integer )

This  boils down to  finding the circumradius of the  circle

We have an inscribed triangle ABC

Semi-perimeter  =  [ 8 + 15 + 12 ]   / 2   =   17.5

Area =  sqrt [ (17.5 ) (17.5 - 8) (17.5 - 15) (17.5 - 12) ]  =  sqrt [ 2285.9375 ] 

Circumradius =   product of the  sides /  4 * Area  =

(8 * 15 * 12)  /(4 sqrt [ 2285.9375] ≈  7.53  =  radius of the  circle

N33E  =  57°

N84E =   6°

S70E  = 340°

4cos 57°  + 5cos 6° + 4.5cos 340°  ≈  11.38

4sin 57° + 5sin 6° + 4.5sin 340°  ≈ 2.34

Direction  =   arctan ( 2.34 / 11.38) ≈ 11.6°  =  N 78.4 E

Magnitude  = sqrt [ 11.38^2 + 2.34^2 ]  ≈  11.61  N

A circle is centered at  The tangent to the circle at  is extended to  Line segment  intersects the circle at  Find the area of the circle.      Measures = OS = 3,   SR = 4,   RQ = 4,  PQ = 8

Draw QS in the direction of S through to M where M  is  on the circle

Secant-Tangent Theorem

PQ^2  = RQ( RQ +RM)

8^2  =  4 ( 4 + RM)

64 / 4  = 4 + RM

16 = 4 + RM

RM = 16 - 4

RM = 12

SM  = RM - SR  = 12 - 4 = 8

Connect OR, OM  =   radius of the circle = r

Note that cos RSO   =  - cos MSO

By the Law of Cosines

r^2  = SR^2 + SO^2 - 2(SR * SO) (-cos MSO)

r^2  = SM^2 + SO^2 - 2(SM * SO) cos (MSO)     simplify this  system

r^2  = 4^2 + 3^2  +2 ( 4 * 3) cos (MSO)     (1)

r^2  = 8^2 + 3^2 - 2(8*3) cos (MSO)

4^2 + 3^2 + 24 cos (MSO)  = 8^2 + 3^2 - 48cos (MSO)

25 + 24cos (MSO) = 73 - 48 cos (MSO)

72 cos MSO = 48 

cos MSO = 48/72  =  2/3      (2)

Sub (2)  into (1)

r^2  = 4^2 + 3^2 + 2 (4 *3)(2/3)

r^2 = 25 + 16

r^2 = 41

Area of circle =  pi * r^2  =  41 pi

Here : https://web2.0calc.com/questions/geometry_55509

Here :  https://web2.0calc.com/questions/polygon_74

Draw PQ , RS

Triangle QVP is similar to Triangle RVS

PQ = sqrt [ QV^2 + PV^2] = sqrt  [ 3^2 + 2^2] = sqrt 13

PQ / PV = RS / VS

sqrt 13  /  2  =  RS / 10

RS =  10 sqrt (13)  /  2  =    5sqrt 13