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what is 20 root 2 minus 20

\(20\sqrt{2}-20=20(\sqrt 2-1)=\color {blue}8.2842712474\)

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(x-4)/6 = (x+5)/5

\(\frac{x-4}{6}=\frac{x+5}{5}\)     Multiply crosswise

\(5x-20=6x+30\)  Change of sides with change of sign

\(5x-6x=30+20\)   add

\(-x=50\)   Multiply on both sides -1

\(x=-50\)

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The nth term is given by :     4 * 5^{n- 1} ...so.... 

4 * 5^{(10 - 1)}  = 4 * 5⁹ = 7,812,500

(x-4)/6    =     (x+5)/5 = -50

              ↑

What is the sign beween the fractions suppposed to be???

value of x 

-7(-4+4)

There is no x.

The term becomes

\(-7(-4+4)=28-28=0\)

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9x + 1/2y = 51   →  18x + y = 102    →  -18x -  y = - 102    (1)

7x + 1/3y = 39   →  21x + y = 117    →   21x + y =    117    (2)

Add (1) to (2)  and we have

3x  = 15    →   x  = 5

So

18(5)  + y  =  102

90 + y  = 102        →  y  = 12

Thanks CPhill!!

y=(2-x)^2+5  →   y=(x-2)^2+5

The vertex is at (2, 5)......the "parent" graph of y = x^2  has its verex at (0, 0)...so the given graph is transformed to the right 2 units and up 5 units

How to solve the equation g(x)=-5 for x if g(x)=-.5x^2+x+4 {nl} Says there's 2 possible solutions

\(-5=-0.5x^2+x+4\)

\(-0.5x^2+x+9=0\)

a        b       c

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)    

\(\large x = {-1 \pm \sqrt{1+4*0.5*9} \over -1}\)

 \(\large x=\frac{-1\pm\sqrt{19}}{-1}\)

\(x_1=1-\sqrt{19}=-3.3588989\)

\(x_2=1+\sqrt{19}=5.3588989\) 

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a. The volume of the composite solid  =  x^3  + 4 * 4 * x  =[ x^3  + 16x^2] inches^3

b.  25x^2  = x^3 +  16x^2    subtract 25x^2 from both sides

0 = x^3  -  9x^2     factor

0  = x^2 (x - 9)     set both factors to 0 and solve for x..... and x  = 0 [reject]  or x = 9 in

Setting the polynomial in part (a) equal to ____  and solving for x results in x=__ , x=__, x=__

                                                                   25x^2                                                0        0         9

- 5  = -.5x^2+x+4   multiply through by -2

10  = x^2 - 2x - 8    subtract 10 from both sides

0 = x^2 - 2x - 18   using the quad formula, the solutions  are

x =  1 - √19     and     x =  1 + √19 

C(16, 4)   = 1820 ways

1. 

inf

∑   2 / [ n ( n + 1) ]     converges to  2   

1

This series converges by a comparison test  to  2 / n^2 [ which conveges by the integral test ]

2 / n^2                       2       1/2     2/9  ..........      

2 / [ n ( n + 1) ]          1       1/3     1/6  ..........

Both series  >  0  for all terms....and, term by term, the second series is smaller than the second

So....if 2/n^2 converges....so does  2 / [ n ( n + 1) ] 

2.

inf

∑   [ n / ( n^2 -  1) ]    

2

This series diverges by a comparison test  to 1 / n  [ which diverges by the p - series test]

1/n                         1/2        1/3      1/4  .........

n / ( n^2 -  1)         2/3        3/8       4/15  ........

Both series  >  0  for all terms....and, term by term, the first series is smaller than the second

So....if 1/n diverges....so does  n / ( n^2 -  1)  

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16

    C

      4                    = (16x15x14x13)  /  (4x3x2x1)

(a) Since AOC is a diameter, then angle  ABC is an inscribed angle that measures (1/2)  of is intercepted arc......and is intercepted arcc = 180°....so  ABC  =  90”

(b) find AC  =  sqrt [ 10 ^2  + 7^2 ]  = sqrt [149]

So .....  7/ sin ACB  = sqrt [149]    →   sin ACB  = 7 / sqrt [149] 

→  arcsin  (7 / sqrt [149]  ) = ACB  = 34.99°  = 35°

(c)  ADB = ACB because they are inscribed angles intercepting the same minor arc,  AB

(d) AD / sin 77  =   7 / sin 35 →    AD  = 7 sin 77 / sin 35   ≈ 11.9

(e)   Angle DAB  =  (180 - 77 - 35)  =  68°

Area  of ABD  ≈  (1/2) (7)(11.9)sin(68) ≈  38.6  cm^2

(f)  ΔDEA ≈ ΔCEB

Area  ΔAED  =  (1/2)(11.9)(DE)sin(35) = 12.3  →  DE = 12.3 / [ (1/2)(11.9)sin(35) ]  ≈ 3.6

So....AD/DE  = BC/CE   →  11.9 / 3.6  = 10/ CE  →  3.6 / 11.9  = CE / 10  →

CE = 10 * 3.6 / 11.9  ≈ 3.03

So area  of ΔBEC  = (1/2)(3.03)(10) sin (35)  ≈  8.69 cm^2

put it into desmos.com

20 in 1 month

20 * 12 = 240 in a year

75=( 2 x 3 x ? ) + 3   subtract 3 from both sdes

72  = 2 * 3 * ?

72  = 6 * ?      divide both sides by 6

12 =  ?

No.......the greatest  factor of a number - in his case, 18 - can never be larger than that number

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