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https://web2.0calc.com/questions/in-triangle-and-let-be-the-incenter-the-incircle

ABCDE = 1200 

1200 prime factorization = 24 * 3 * 52

Rolls: 5, 5, _, _, _

Fit 24 and 3 as product of 3 numbers (6 or less):

4, 4, 3

6, 4, 2

5, 5, 4, 4, 3 and 5, 5, 6, 4, 2

Ways to arrange 55443 = 5! / (2! * 2!) = 30

Ways to arrange 55642 = 5! / 2! = 60

30 + 60 = 90

Graphing the equation on a ti or Desmos is simplest. to the nearest tenth, it is 3.9

h(d) = 0 when he lands

0 = -3.9d^2 + 13 + 8.7        Quadratic Formula shows d = 3.9 meters

solution is from user melody https://web2.0calc.com/questions/counting_88

1 apple, 3 peaches     3*5C3

2 apples, 2 peaches     3*5C2

3 apples 1 peach       1*5

Then add them up.

Slope of the line is (17 - 11) / (-2 --5) = 6/3 = 2. Find the y intercept (b) is 21. y = 2x + 21

 We  need  

(3x  +  m)  (3x + m )   =  9x^2  +  6mx  + m^2

So

6m  = 24

m  =4

So

(3x  + 4)^2   =  9x^2  + 24x  +  16

m^2  =  16     =  a

a = 16

Triangle angle bisector theorem:

An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.

24 is to 28    as    AX is to 21

24/28 = AX/21

21 * 24/28   = AX  

Angle COA  =  360  - 2 (100)   =  160

Angle CAO =   (180  -160)  /2   =   10°

And since   BO  = AO  then  angle BAO  =  (180 - 100)  / 2     =  40

So....angle  BAC  =   angle COA  +  angle BAO   =  10  + 40   =  50°

A

6                  10

B            8                   C

This  is a 6 - 8 -10    right  triange

Since  AD is  an angle  bisector   we have  that

AD / DC  =   AB  / AC

AD /DC  =  6 / 10

AD/ DC  = 3/5

AD  =  3/ ( 3 + 5)  * 8     =   (3/8) (8) =  3

AD^2  =  3^2   =   9  

Let XY be a tangent to a circle, and let XBA be a secant of the circle, as shown below. If AX = 15 and XY = 8, then what is AB?

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AX = 15         XY = 8         BX = x

x(AX) = (XY)²          x = 4 ⁴/₁₅

AB = AX - x = 10 ¹¹/₁₅   

a) for REAL roots, the discriminant of the quadratic equation must be > 0

         a = 1      b  = (3k-1)        c  =10 + 2k

b^2 - 4ac >0      sub in the values above and solve for 'k'     .... can you take it from here with that help?

Second part    For TWO EQUAL roots the discriminat must = 0

b^2 - 4ac = 0          a   b   c  are the same as above

We  have  this relationship

XY^2   =   XB  * AX

 8^2  =   XB  (15) 

64  =  XB (15)

64 / 15   = XB

AB =  AX  - XB   =    15   -  64/15  =    (225 - 64)  / 15      =    161/15 

The area of an isosceles trapezoid ABCD is 24 cm^2, and the diameter of an inscribed circle is 4 cm. Find the perimeter of the trapezoid.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

[ABCD] = 24           r = 2               Circle center => O         ∠AOD = 90º

AD = BC = (1/2[ABCD]) / r = 6

AD + BC = 12

AD + BC = AB + CD

So, the perimeter is 24 cm

Here is a similar question:  https://web2.0calc.com/questions/power-of-a-point-question 

Second one

-4.9t^2  +  20t  +  6

The  x coordinate of  the vertex =   -20 / ( 2 * -4.9)  =   100/49  sec

The  max  ht = y coordinate of the vertex  =    -4.9(100/49)^2  + 20(100/49)  + 6     ≈   26.4   m

First one

x^2  - 4x  - 1  =  2kx   - 10k

x^2  - (4 + 2k)x  + (10k - 1)   =   0

If we  have two equal roots......then  the  discriminant = 0   ......so.....

(4 + 2k)^2  - 4 (1) (10k - 1)  =  0      simplify

4k^2  + 16k + 16  - 40k  + 4  =  0

4k^2  - 24k +  20  =   0         divide through  by  4

k^2  - 6k +  5  =  0

(k - 5)  (k -1)   = 0

Set  each factor to 0  and solve for k   and   we get  that

k=  1      or  k =  5

no problem 

Correct! Thank you!!!

We  have  that

y =  a ( x + 2)  ( x + 3)

And since it passes  through  (5, 10)    we can find "a"  thusly

10  = a ( 5+2) ( 5 + 3)

10  = a  (7)(8)

10  = a  (56)

a = ( 10 / 56)  =   5 / 28

Then we  have  that

y =  (5/28) ( x + 2) ( x + 3)

y = (5/28)  (x^2  + 5x + 6)

y = (5/28)x^2  + (25/28)x  + 15/14

x + y = -10 and xy = -11, from the second equation we get, y = -11/x. Sub that into the first equation, we get x - 11/x = -10. Can you take it from here?

Cubes that have exactly 1 black face would be 6 * 3 * 3 = 54

Assume  that   the  child sits in seat   1

Count  the number of ways   that  the  child  CANNOT  sit next to  either  parent

They   must  sit in   seats   3, 4      3,5   or    4,5  

They  can sit  2 ways  in either seats 3,4 , 3,5  or  4,5   and for each of these, the  other people (not counting the  child) can be arranged in  3! ways  =       3 (2)(3!)   =  36  ways

Total arrangements  =  5!  =  120

So.....the  ways  the child CAN sit next to at lrast one of the parets =  120  - 36  =   84

{ The Guest  and I agree  !!!  }

Jonathan and Kenneth shared a sum of money in the ratio of 3 : 4. After Jonathan spent 1/6 of his money and Kenneth spent $15 of his money, both Jonathan and Kenneth had the same amount of money left. How much money did they have at first!

J : K = 3x : 4x

J spent 3x/6

K spent $15

3x - (3x/6) = 4x - 15

x = 10

Jonathan had $30

Kenneth had $40

i'm assuming there are 2 parents... Answer should be 5! - 3!3! = 120 - 36 = 84...correct me if im wrong