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I will give you 2 good hints.

You can come back with your thoughts if you need more help.

1)   let y=3^x

2)  \((a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4\)

We    have  the form

y = ax^2  + bx  +  c

Since   f(4)  = 0   then  4 is a root

And   f(1)  =  -24

And f(7)  = 60

So    we  have  this system of  equations

a(4)^2  + b(4)  + c   = 0

a(1)^2  + b(1) + c  = -24

a(7)^2 + b(7) + c  =   60         simplify

16a  + 4b +  c  =   0             (1)

a    +   b    +  c  =  -24          (2)

49a  + 7b + c  =  60             (3)

Subtract  (2) from (1)

15a + 3b  =  24  (divide through  by 3) ⇒   5a + b  = 8       ⇒  b = 8 - 5a    (4)

Subtract  (1)  from ( 3)

33a + 3b  =  60  (divide through by 3) ⇒ 11a + b  = 20         (5)

Sub (4) into (5)

11a  + (8 - 5a)  =  20

6a  =  12

a = 2

b = 8 - 5(2)  =  -2

a + b + c  = -24

2  - 2  + c =  -24

c = -24

So  our polynomial  is

f(x)   = 2x^2  - 2x  -  24

f(-1)  = 2(-1)^2 - 2(-1)  - 24   =   -20

Here's a grqph  : https://www.desmos.com/calculator/xzq6el9uj2

sin^2x  +  3cos x  -  2   =  0

( 1 - cos^2x)  + 3cosx  -  2  =  0

-cos^2x + 3cos x - 1 =  0

cos^2 x - 3cosx + 1   =  0                 let cos x =  a

a^2  - 3a   =   -1          completete the square  on a

a^2  - 3a    +    9/4  =  -1  + 9/4

(a - 3/2)^2  =  5/4      take both roots

a - 3/2   =  sqrt (5)/2              a  - 3/2  =  -sqrt (5)/2

a   =  sqrt (5)/2 + 3/2                         a  =  3/2  - sqrt (5)/2

cos x  = sqrt (5)/2 + 3/2                  cos x =  (3 - sqrt 5)  / 2

Not possible

cos^3x  + sec^3x   =

(cos x  + sec x)  ( cos^2 x  - cosxsecx  + sec^2x) =

(cos x + secx) ( cos^2 x  + sec^2x  - 1)

[ (3 - sqrt 5)/2  +   2/ (3 -sqrt 5) ]    [  ( ( 3 -sqrt 5)/2 )^2   +  (2/(3 -sqrt 2))^2   - 1 ]   =  18

Let p be the cost of each pendant.

Since there are two necklaces, we can subtract 2(3.50) from 13.80 (assuming that 3.50 is the cost of all of the beads per necklace) - which gives us 6.80.

Since there are two pendants, we divide 6.80 by 2 which gives us 3.40 as the cost for each pendant.

Hope this helped!

Caffeine :)

Angles   30º     45º       side    √3+1 cm

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

With this information, 3 similar triangles but different in size can be made.

Which one do you want? 

Already posted: https://web2.0calc.com/questions/confusing-algebra

Good job!!!

a = 15        b = 20

Let E  be  the intersection  of  BC with the  circle

By  the secant tangent theorem,   

BE (  BE + EC)  = AB^2

BE  ( 8) =  6^2

BE (8)  =  36

BE  = 36/8   =  18/4  =  4.5

So   chord CE  = 8  - 4.5  =  3.5

If  we call the midpoint of this chord, F  and we  draw   OF....then triangle    ODF is right with OD the hypotenuse

And  DF  =  10 + CF =  10  + CE/2  =  10 + 3.5/2   =   11.75

And by  the  Pythagorean Theorem,   OF^2  = OD^2  -  DF^2  = 5.9375

And we can  draw another  right triangle  OFC  with  the radius  of the  circle, r, as a hypotenuse, legs CF  and  OF

So

r^2  =  CF^2 + OF^2

r^2  =  (1.75)^2  + 5.9375

r^2   =  9

r =  3

If BD is 17 then b or d is 17, but d cannot be 17 otherwise it would be fractions. therefore, d=1 and a=6 and b=17 so c=15/17 so 6*15/17=90/17

The center of the ellipse is (0,0)  =  O 

1/2 of the  chord length  will be the  x value associated  with one of the other vertexes  on the positive side of the  y axis      ......the other vertex will  have an x coordinate of - sqrt(7)

So.....to  find the  y value   we have

(sqrt 7)^2  / 16 +  y^2/9  = 1

7/16  +  y^2/9  =  1

y^2/9 = 1 - 7/16

y^2/9  =  9/16       multiply  both sides by 9

y^2 =  81/16      take the positive root

y=  9/4  =    the  height of the triangle

The  area of the triangle =    (1/2) chord length  * height  =    sqrt (7) *(9/4)  =   2.25 sqrt (7)  units^2

You're on the right track....we have a pattern that you found as

Term    1   2   3    4   5   6   7    8    9  10   11  12   13    14  15  16

           3, 27, 24, 3, 21, 18, 3, 15, 12, 3,    9,   6,     3,   [3,    0,   3] , [ 3, 0, 3], ]3, 0 , 3]

Notice that......starting with the  14th term   we  have  the repeating pattern   3,0,3

Notice that   after  this    ( term   number + 2)   ] / 3       gives us  what we need  to determine the 2016th term

Notice  that

(14 + 2) / 3  =  16/3 =    5  +   1/3    =    1st   number  in the pattern  =  3

(15 + 2) / 3  =   17/3  =  5   +  2/3  =       2nd number in the pattern  =  0

(16 + 2) / 3  = 18/3  =    6        whole number   result gives us the 3rd number in the pattern   = 3

So....to determine the  2016th   term we have

(2016 + 2 ]   / 3   =  2018 / 3  =    =   672  + 2/3   =  2nd number in pattern  =   0

If perimeter is 24 then   PS and QR are 7   (7+7+5+5)  = 24

   height  ST must be   4     because   area = 7 x4 = 28

      then RT = 3    (3-4-5 triangle  via Pythag theorem)

then QOST = 4 + 3 +7 + 4 +3+7 = 28 units

Ok, thanks :D

See  here, piano123  :

https://web2.0calc.com/questions/solve-this-with-good-explaination-please

Call   a, b  the legs  and the  hypotenuse, c

a + b +  c   = 60

a +  b =   60 - c                square both sides

a^2  + 2ab + b^2   =   3600 - 120c + c^2

Note   a^2 + b^2  = c^2

So....we have that

2ab  = 3600  - 120c

ab  =  1800 - 60c     (1)

Area of  triangle   = (1/2)  ab

Also

Area =   (1/2)c *12       so

(1/2)ab = (1/2) 12c

ab  =  12c        (2)        sub  (1)  into (2)

1800  - 60 c  =  12c

1800  =  72c

c = 25

a + b =   60  - c

a + b  =  60 -  25

a + b  =  35  =   sum of the leg lengths

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Also, please include equations not just the answer. Thx :)

Assuming   x-5y = -7

Slope of first line =   1/5     perpindicular = -5

   Py = -15x + Q

     y = - 15/P  + Q /P       -15/P = -5     so P = 3

15(8) + 3(3) = Q      Q = 129

2^x   =  9

27^y  =   2

Rewrite  as

x =   log 9/log 2 =     log 3^2  /  log 2   =    2log 3  / log 2

y =  log 2 / log 27      =    log 2 / log 3^3  =   log 2  / {3 log 3 ]

So

xy  =        2log 3 / log 2    *   log 2  / [ 3 log 3 ]    =   2/3

Maybe you meant   x - 5y  =  -7     ???

If so....rearrange as

-5y  =  -x - 7          divide  both sides by   -5

y = (1/5) + 7/5

The slope of this line   =   1/5

A perpendicular line has the slope   -  5

Rearrange     15x + Py   = Q     as

Py   =   -15x  +   Q         

y  =   (-15/P)x  +   Q/P     

To  have a slope of  -5,  P  must  be  =  3 

So   we have     

y =  -5x  +  Q/3

And since  (8,3)   is on this line we have that

3 = -5(8)  + Q/3

3   = -40  +  Q/3

43  =  Q/3          multiply through by 3

129 =   Q

So   (P,Q)   =   (3, 129)

....is that supposed to be    x - 5y = - 7 ??

12 x 12 = 144

This question (or similar) was posted the other day....with the same mistake....   8,3 is not on the line    x - 5y = 7