All Answers 358087 Answers

ab^4 = 48 a^2 b    <==== solve this for 'a'

b^3 = 48a 

a = b^3/48         < sub this definition of 'a' into the second equation.....then solve for 'b'

a^2 b^5 = 6a^3 b       <==== it will be easier if you simplify this first

b^4 = 6 a 

Now sub in the value of a 

b^4 = 6 (b^3/48)      and solve for 'b'

b = 6/48 

b = 1/8         

......

I see you wonder about  the points associated with your account....

   I believe you get a 'point'  when you post a question .....and a 'point' when you answer another poster's question......you can 'upvote' yourself for 1 'point' too when you answer a question......other members/posters can also 'upvote' your answer ( well......that was the way it used to be......maybe no longer) .

  I might get a point for this post, but I will not upvote myself for it.....but you can see if YOU can upvote it if you like..... to answer your Q

The number of days or hours you spend on the site  is irrelevant.....

~EP

g = grogg age now     d = dad's age now 

(d-7) = 12 ( g-7)       <===== given in first sentence

d-7 = 12g - 84 

d = 12g - 77   

and 

(d-3) = 7 (g-3)     <===== given in second sentence 

d-3 = 7g -21 

d = 7g - 18                   Now you have TWO definitions for what d is equal to .....equate the two 

12g -77 = 7g - 18 

5g = 59 

g = 11  4/5 y/o      then    d = 12g-77   shows d = 64  3/5  y/ o 

\(\frac{3-z}{z+1} \ge 2 + \frac{z}{z-7} \)

Rewrite  using x for z

[3 - x ] / [ x + 1 ] - 2 -[ x ]  / [ x - 7]   ≥  0     

Multiply through by ( x + 1) ( x -7)     and  make this  an equality

(3 - x) (x - 7)   - 2 (x + 1)(x -7) - x ( x + 1)  =  0

-x^2+ 10x  - 21  - 2 (x^2 - 6x - 7)  - x^2 - x   =  0      

-x^2 + 10x - 21  - 2x^2 + 12x + 14 - x^2 - x =  0

-4x^2 + 21x  - 7  =   0

Using the Q Formula

x =  [ -21 ± sqrt [ 21^2  - 4(-4)(-7) ] ] / -8

x = [-21 + sqrt [ 329 ]  ] / -8   = [21 - sqrt [329] ]/ 8     ≈  .358     

or   

x =  [ -21 - sqrt [329] [ / -8  =   [ 21 + sqrt [ 329] ] /  8 ≈   4.89

We  have these intervals to consider  ( x not defined at -1, 7 )

Interval                  Test Point        Test point makes the original inequality

(-inf , -1)                    x =  -2                       False

( - 1 , .358]                x = 0                          True

(.358  , 4.89)             x = 1                         False

[ 4.89 , 7 )                 x = 5                          True

(7, inf)                       x = 8                          False

z =  (-1 , [21 - sqrt [329] ] / 8 ]   U  [  [21 + sqrt [ 329] ] / 8 , 7 )

4t^2  ≤ -9t + 12 - 14t + 28      simplify as

4t^2 + 23t - 40   ≤  0          (1)            set  as an equality

4t^2 + 23t  -40  =  0

Using the Q Formula

t =    [  -23   ± sqrt ( 23^2  - 4 * 4 * -40 )  / 8

t = [ - 23 ± sqrt [ 1169 ]  / 8

t  ≈  - 7.14      or    t  ≈  13.99

0  lies in the  interval between these two  values......if t = 0 , then (1)  is  true 

Then  t  = (  [-23 - sqrt (1169) ] / 8   ,   [ -23 + sqrt (1169) ] / 8  )

Thanx, Chris !   

     I was wondering if I had this correct.....now I know it likely is correct.....or maybe we BOTH got it wrong ! HaHa 

~EP

x(x + 6) > 16 - 7x + 24  simplify 

x^2 + 6x > 40 - 7x 

x^2 +13x -40 > 0 

(x-10)(x-4) > 0      shows the intervals    - inf to 4    4 to 10    and  10 to + inf 

Subbing in values in these intervals will show that only the  4  thru 10 interval is not a solution   and 4 and 10 are not solutions

(-inf, 4)  U  ( 10, + inf) 

x^2 + 4x -12 le  3x + 7

x^2 + x -19 le 0      set up as an equality  , complete the square on x

x^2 + x  + 1/4  =   19 + 1/4

(x + 1/2)^2  =  77/4     take both roots

x =  [ -1 - sqrt 77 ]  /  2   ≈   -4.89    or        x =  [ -1+ sqrt 77 ] / 2    ≈ 3.89

Test 0 in  the original inequaity  and it makes it true

Then  the  answer  comes from  the interval   ( [ -1 - sqrt(77) ] / 2 , [ -1 + sqrt (77) ]  /2 )

n^2 < 144 + 40n

n^2 - 40n  - 144   < 0

Make this  an equality   and  complete the square on n

n^2  - 40n   = 144

n^2  -40n + 400  = 144 + 400

(n - 20)^2 =  544       take  both roots

n - 20  = -sqrt (544)

n = 20  - sqrt (544)           and      n=   20 +  sqrt (544)

n ≈ -3.32                                       n ≈  43.32

The   answer  either comes from  the interval between these two  numbers or from  the two intervals on  either side of these two numbers

n = 0   is in the first interval and   it solves the inequality  because 0 < 144 + 40(0) is  true

So

The answer is   20 -sqrt 544 < n < 20 + sqrt 544

I don't find an easier way, either, EP.

I did it with just some  Geometry

Here's my solution  (but it  might not  be  any  easier)

Like EP, I found BT to be 30

We can find the coordinates of T  by finding the intersection of two circles  (this is the key !!!)

One with a center of (0,30) and a  radius of 30

One with a center of (15,0) and a  radius of 15

The equations are

x^2  + ( y - 30)^2  = 900    →  x^2 + y^2 - 60y   =  0   (1)

(x - 15)^2 + y^2  = 225     →  x^2 - 30x + y^2  = 0     (2)

Subtract these

-60y + 30x  =  0

x = 2y

Sub  this into (2)

4y^2 - 60y + y^2  = 0

5y^2 -60y  =  0

y - 12  = 0

y =12

x = 2(12)   = 24

T = (24,12)

Slope   of   line through BP = ( 30 - 12) / ( 0 - 24) = -3/4

Equation of this  line  is  y =(-3/4)x + 30

Now, as EP  said, put this into  the  equation of the circle  x^2 + y^2  =900  and find the x coordinateof P  as  28.8

And  -(3/4)(28.8) + 30  =   8.4 = the y coordinate of P

TP = sqrt [ (28.8 -24)^2  + (12 - 8.4)^2 ]  = sqrt [ 4.8^2  + 3.6^2 ]  =  sqrt (36)   = 6

Nice job, EP!!!

Much easier if you notice that this is a quadratic X^2 + X - 12= 0, where X = (4x -16)/(3x - 4).

Solve for X (= -4 or 3 ) and then for x.

\($x \mathbin{\spadesuit} y = \frac{x^2}{y}$ for all $x$ and $y$ such that $y\neq 0$ \)

\(Find all values of $a$ such that $a \mathbin{\spadesuit} (a + 1) = 9$. \)

a (spade) (a + 1)  =       a^2 / (a + 1)   =  9

a^2 = 9(a + 1)

a^2  = 9a + 9

a^2  - 9a   - 9    =  0

a^2 - 9a =   9             complete the square on  a

a^2  - 9a  + 81/4  =  9 + 81/4

(a -  9/2)^2  =  117  / 4             take  both roots

a - 9/2  = sqrt (117) / 2

a = [ 9 + sqrt (117) ] / 2         or        a =   [ 9 - sqrt (117) ]  / 2

Not possible

If the axis  of symmetry is x  =1 , then the parabola   turns  either upward or  downward

If the x intercept is at 0, then y at that point  also = 0  

Then  we can't also have a y intercept at -1 because this parabola would fail the vertical line test  because it would have two y intercepts  at (0, -1)  and (0, 0)

Simplify as

8x^2 - 11x - 45 =  0 

Quadratic Formula

x =  [ 11 ±  sqrt  [ 11^2  - 4 (8) (-45) ] ] /  [2 * 8 ]  = 

[ 11 ±  sqrt [  121 + 1440 ] ]  / [16 ]  =

[ 11± sqrt  [ 1561 ] ]  /  16  

13.009 + 12.83 is equal to 25.839.

Answer:  25.839

39 x 429 / 2 = 39 x 214.5 = 8365.5

Answer: 8365.5

Changing angle XAY to  16°

Then angle  AYX  = angle  AYB =  measure (AXY) - measure (XAY ) =    90°  - 16°  =  74°

The answer I get is 36 cm ...... It took a while for me to figure a way to get the answer.....there is likely an easier way but here is my (long -hopefully-correct) solution:

See image :

Use arc tan to find the 63.435 angles then find the 53.13 angle     ( 63.435+63.435 + 53.13 = 180 ° ) 

           then use sin and cos functions for angle 53.13 °  to find the coordinates of T

              then use the coordinates of B and T to find the equation of the line BP to be   y = - 3/4 x + 30 

                   the equation of the Quarter circle is    x^2 + y^2 = 900    or    y = sqrt(900-x^2) 

                         equate this to the line equation and solve for the 'x' coordinate of  point P     then use the line equation and this value of the 'x' coordinate  to find the value of the 'y' coordinate for point P

           Now you have the coordiantes of the two endpoints of the chord line 

                        B    and   P  ====>    then finish by using the distance formula to find  BP = 36 cm 

There really must be an easier way !!!    Whew !   

J /  S = 3/2

J =  (3/2)S

( J + 12 )  / ( S  - 1)  =   2/1

((3/2)S+ 12) /  (  S  -- 1)    =  2/1               cross-multiply

(3/2)S + 12  = 2 (S - 1)

(3/2)S + 12  = 2S   - 2

12 + 2 =  2S - (3/2)S

14 =  (1/2)S

14 ( 2/1)  = S

28  = S

J = (3/2)(28)  =  42

After changes    J + 12 =  54           S - 1 =  27

54 +  27   =    

81 total  students

When the perimeter is doubled  the scale factor  =  2

Original area  =   New Area /  Scale Factor^2  =   3600 / 2^2  =  3600 / 4 =  900

ACB is  an inscribed angle in the  circle  which intercepts the same  arc as central angle AOB....ACB has half the measure of  AOB

So

ACB = (1/2)AOB

2 * ACB = AOB

2 * 40  = AOB  =  80 (degrees)