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cube root of y^3 = y

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8x^2 -3 = -155
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8x^2 = -152
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x^2 = -152/8 = - 19
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taking sqrt of both sides
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x= sqrt(-19) = 4.36 i,,,,,,,,where i = sqrt(-1)

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x= 6.017,,,,,by trial and error,,,assume x,,,calc eqn
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check,,,6.017^2 +2^6.017=100.963 = 101

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(x+2)(x-3)(x+4)
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use FOIL first
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(x+2) ( x^2 +4x -3x -12)
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(x+2) ( x^2 + 1x -12)
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now use distribution
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x(x^2 +x -12) +2 ( x^2 +x -12)
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x^3 +x^2 -12x +2x^2 +2x -24
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x^3 +3x^2 -10x -24
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checking,,,if x=0,,, original and answer = -24,,ok
if x=1,,,both = -30,,,ok

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v=d/t
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(1) v=d/5
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(2) 1.2 v = d/t
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from (2),,v=d/(1.2 t) = d /5,,,,,,from(1)
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1.2 t =5
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t= 5/1.2 = 4.17 min
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checking,,,,let d=10, v= 10/5 =2
v2= 10/4.17 = 2.4
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2.4 / 2 = 1.2,,,or 20 % higher,,,ok

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congruent

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congruent

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852 <= v <= 1153

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Let s = Sanjay age,,,b= Brother's age
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(1) s = (1/2) b
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(2) ( s+6) = (3/5) (b+6)
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subst (1) in (2)
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(1/2) b +6 = (3/5)b + 18/5
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6- 18/5 = (3/5 -1/2) b
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2.4 = .1 b
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24 = b,,,,,,s= 1/2 b = 1/2 *24 = 12
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check
(1) 12 = 1/2 * 24 ,,,,ok
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)2) 12 +6 = (3/5) ( 24 + 6 )
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18 = 3/5 *30 =18 ,,,,,,ok

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A + B = 180,,,,def supplementary
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B= (70-x),,,,given
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A +(70-x) =180
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A + (70-x) = 180
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A = 180 -70 +x
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A= 110 +x
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check, x=10, (70-x) =60,,,A=120,,,answer,,,110+10=120,,,,ok

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sqrt 56ab^3 / sqrt 7a
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sqrt{ 56ab^3 / 7a }
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sqrt { 8b^3)
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sqrt{ 4 *2 *b^2 *b }
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2 b sqrt( 2b)
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check let b=1,,both answer and original = 2.828,,,,ok

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(3x) /(7-x) > x
3x > x(7-x)
3x> 7x -x^2
0> -x^2 +4x
0> x(-x +4)
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x<0,,,,x>4
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check ,,x<0,,use x=-1, -3/8 > -1,,,,,ok
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0less than x less than 4,,use x= 1,,,3/6>1,,,,,,not true
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x>4,,,use x=5,,,15/2 > 5,,,,,ok
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therefore ,,,x<0,,,x>4
.

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(-7x+5) / (6x^2 +3) >0
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(-7x+5) >0
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5> 7x
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5/7 >x

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Good work,,,,but watch signs for last 2 steps
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log{ (1-x) / (x+2) } = log (x^2)
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(1-x) / (x+2) = x^2
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(1-x) = (x^2) (x+2)
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1-x = x^3 + 2x^2
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0 = x^3 +2x^2 +x -1

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4.75 gal / 1.25 hr = 3.5 gal /hr

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3^w * 3^5^-w *3^1
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3^w * 3^(-5w) *3^1
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3^(w -5w +1)
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3^(-4w+1)
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check, w=o, original and answer = 3
w=1,,both = 1/3^3
w=2,,both = 1/3^7,,,,ok

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(1) diagon separates a square into 2 equal triangles. legs = "s",,
diagonal =d,,,,as it is a right triangle(square corner),,
use pythagorous,,c^2 = a^2 +b^2,,,with a = b =s,,,c=d,
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d^2 = s^2 +s^2 =2s^2,,,,d = sqrt 2 s^2 = s*sqrt2,,,,,
.
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or (2) recognize that the triangles are ,,,45-45-90, with side ratios,
1-1-sqrt2,,or s-s-s*sqrt2,,,and see d= s*sqrt2

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(1) x =y-6,,,,,,x = smaller, y = larger
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(2) xy = 48 + x^2,,,,,product means multiply
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subst (1) in (2)
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y(y-6) = 48 + (y-6)^2
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y^2 -6y = 48 + y^2 -12y +36
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6y = 84
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y = 14
.
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x=y-6 = 14-6 = 8
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check in (2),,14 * 8 = 48 + 64 = 112,,,,ok

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f(x) = 1/(4-8x)
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f(r-3) = 1/ { 4-8(r-3) }
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f(r-3 = 1/ (4-8r+24)
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f(r-3) = 1/ (28 -8r),,,,,,or B

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In solid figures, if linear dims are proportional, figures are similar.
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Scale factor by radius are,,4/8 = 1/2,,,by height,,10/20 =1/2,,,,
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Therefore they are similar

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x + 1/x =8
(x^2 +1 ) / x =8
x^2 +1 = 8x
x^2 -8x +1 =0
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using quadratic eqn,, x= 7.873,,.127
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substitute either in
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X^4 +(1/x^4) = 3842
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checking answers, this matches (E) (8^4 -2^8 +2)

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(x-3)^2 +(y+4)^2 =25
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Remember form is,,,,(x-h)^2 +(y-k)^2 = r^2,,,(h,k) is center
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in our case, h=3,,,k=-4,,,r=5,,,or center is (3,-4)

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y= x^3 -15x^2 +12x +6
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y' = 3x^2 -30x +12
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to find vertex, set slope(y') =0
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3x^2 -30x +12 =0
3(x^2 -10x +4) =0
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solve using quadratic eqn,,,x= 9.58,,,.417
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to find y,,,substitute in original eqn,,,,(9.58,-376.5),,,(.417,,8.47)
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since this is a positive cubic,,rises to right,,
1st vertex,,,(.417,8.47) is concave down,,,and
second vertex (9.58, -376.5) is concave up
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checking,,,(.5,8.375) (.3,8.277),,,ok cc down
(9.5,-376.4),(9.7, -376.3) ,,,,ok cc up

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Geometric series has form,,,,a(n) = a(1) * r^(n-1)
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a(1) =5,,,,r=4.2,,,,,n= 6
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a(6) = 5*4.2^ 5 = 6534.56
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checking,,5,,21,,,88.2,,370.44,,1555.848,,6534.56,,,,,ok

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Please see problem # 201610,,,Thanks

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Geometric sequence has form,,,,a(n) = a(1) r^(n-1)
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given,,,a(1) =1,,,r=1.5
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a(n) = 1* 1.5^(n-1)
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for n=9,,,,,a(9) = 1*1.5^8 = 25.629
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checking,, 1,,1.5,,2.25,,,3.375,,5.0625,,7.59375,,11.390,,17.086,,25.629,,,,,ok

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Let,,w,x,y,z,,,be four digits in order.
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x=1/4 y
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y=2w,,,,therefore x = 1/4 * 2w = w/2
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z=w
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rewriting the digits,,,,,w,,w/2,,,2w,,,w
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2w is "largest" digit,,,,like to set to 9 but also have a digit, (w/2) ,,,therefore 2w =8 is largest possible.
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restating digits,,,4,,,2,,,8,,,4,
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checking, numbers match criteria.

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(1) d= (r-6)3
(2) d= (r+6)2
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3(r-6) = 2(r+6)
3r-18 = 2r +12
r=30
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check,,d = (30-6)3 =24 *3 = 72
d= (30 +6) 2 = 36 *2 = 72,,,,ok