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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

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Remember alogb= logb^a,,,,and log a - log b = log (a/b)
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5 log (3x-11) - 7 log(x )
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log (3x-11)^5 - log (x)^7
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log { (3x-11)^5 / (x)^7 }
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check,,,,let x=5, both original and answer = -1.882

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we believe the numerator of second multiplier should be (21) not (2)
Please confirm
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{ (x^2 +x-2) / (x^2 -7x) } * { (5x^2 -38x +21) / (x^2 +2x) }
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factor
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{ (x-1)(x+2)/ (x)(x-7)} * { (5x-3)(x-7) / x(x+2) }
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simply by canceling opposing like terms
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{ (x-1)(5x-3) } / x^2
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check by allowing x=1,,both original and answer are zero
try allowing x=2,, both are 1.75,,,,ok

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( 2x) / (2x-3) = ? / ( 8x^3 -27)
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factor second den,,which is a difference of cubes
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(2x) / (2x-3) = ? / (2x-3){4x^2 +2x(3) +9 }
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2x /1 = ? / ( 4x^2 +6x +9)
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cross multiply
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? = 2x(4x^2 +6x +9)
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? = 8x^3 +12x^2 +18x
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check for x=1,,both original and answer has ,,? =38,,,ok

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As a linear relationship, it follows the form, y=mx +b.
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In this case,,,R=mt +b,,,with points (0,46.4),,and (70,45)
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m= (y2-y1) / (x2-x1) = - (46.4 -45) /(1990 - 1920) = -1.4 / 70 = -.02
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R = (-.02)t +b,,,,,@(0,46.4),,,46.4 = 0 +b,,,b=46.4
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Base eqn ,,,,R = -.02t +46.4
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@ 1990, t=1990-1920=70,,,R=-.02(70) +46.4 = 45,,,,,good check
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@ 2003, t=2003-1920=83,,,R=-.02(83)+46.4 = 44.74
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@ 2006, t=2006-1920=86,,,R=-.02(86)+46.4 = 44.68
.
Have a great day

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25 * 17
.
12,,,,34,,,,,even(12), not add
6,,,,,68,,,,,even (6), not add
3,,,,,136
1,,,,,272
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,,,,,,408 (136 + 272) ,,,,,close to 425
.
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48 * 39
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24,,,,,78, even(24), not add
12,,,,,156, even(12), not add
6,,,,,,312, even(6), not add
3,,,,,,624,
1,,,,,,1248
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,,,,,,,1872,,(624 +1248),,,,exactly 1872
.
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120 * 42
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60,,,,,,84, even (60), not add
30,,,,,168, even(30), not add
15.,,,,336
7,,,,,,672
3,,,,,1344
1,,,,,2688
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,,,,,,5040,,,,(336+672+1344+2688),,,exactly 5040
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again, first column divide by 2,(drop remainders),,second column multiply by two. Cross out even numbers in first column , and mates in second column. Then add remaining second column.

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L= 2C +1.57(D+d) + (D+d) / (4C)
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Please check to assure this is the problem
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substituting, C=19, D= 8,,d= 5
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L = 2(19) +1.57{ (8) +(5) } + { (8) + (5) } / { 4(19) }
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L= 38 +1.57(13) + (13)/ (76)
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L= 38 + 20.41 + .1711
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L = 58.58

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Let x .
x +2y =45
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x,,,,,,,y
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1,,,,,,,22
3,,,,,,,21
5,,,,,,,20
7,,,,,,,19
9,,,,,,,18
11,,,,,,17
13,,,,,,16
15,,,,,,15,,not true as x not less than y
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These are just positive integers. Negative integers, and decimals add many more possible combinations.
.
Perhaps the problem has other limitations ??

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1) to construct a new equilateral triangle of side "s", start with the existing equilateral triangle, Extend 2 sides until they are "s" long, connect these points.
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2)altitude is perpendicular distance from the vertex to the opposite side
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Median is the distance from the vertex to the opposite side
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They are equal in isosceles(when base is odd length) or equilateral triangles
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3)Construction easily makes 180 degree and 90 degree angles.
bisection or repeated bisection yields, 45, 22.5, 11.25, etc angles
Adding or subtracting from each other yields combinations.
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a 30-60-90 triangle has sides of 1-2-sqrt3. Constructing a 30 degree or 60 degree is readily done using length ratios of 1 &2. Bisecting these repeatedly yields 15, 7.5, 3.75, etc. degrees. These can be added or subtracted to find numerous angles.

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2y -3/2 = sqrt 729 /(4+y)
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(2y-1.5)(y+4) = 27
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2y^2 +8y -1.5y -6 -27 =0
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2y^2 +6.5y -33 =0
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Use quadratic eqn to solve
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a= 2,,,,,b= 6.5,,,,,c=-33
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y=[ -(6.5) +/- sqrt{ (6.5)^2 -4 (2)(-33) } ] /2(2)
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y= [ -6.5 +/- sqrt{ 306.25} ] /4
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y= [ -6.5 +/- 17.5] / 4
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y = -6,,+ 2.75
.
check by inserting in original eqn,,,,,ok

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( x^4 -2x^2 +3) / ( x+1)
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.,,,,,,,,,,,,,,,,,,,x^3 -x^2 -x +1 +(2)/(x+1)
,,,,,,,,,,,,______________________________
,,,,,,x+1,,!,,x^4 +0x^3 -2x^2 +0x +3
.,,,,,,,,,,,,,x^4 +x^3
,,,,,,,,,,,,,,________
,,,,,,,,,,,,,,,,,,,-x^3 -2x^2
,,,,,,,,,,,,,,,,,,,-x^3 -1x^2
,,,,,,,,,,,,,,,,,,,__________
,,,,,,,,,,,,,,,,,,,,,,,,,-x^2 +0x
,,,,,,,,,,,,,,,,,,,,,,,,,-x^2 -1x
,,,,,,,,,,,,,,,,,,,,,,,,,________
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,+1x +3
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,+1x +1
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,_____
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,+2
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Polynomial division is similar to regular long division,,
divide, multiply, subtract, bring down,,,,,,repeat
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To check, multiply quotient by divisor aand add remainder
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(x^3 -x^2 -x +1) * (x+1) = x^4 -2x^2 +1
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adding remainder(+2) = x^4 -2x^2 +3,,,,,,ok






.
.

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(c+4)(c-2)(c-7) > 0
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lets solve as equality first
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(c+4)(c-2)(c-7) =0
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c+4=0,,,,c=-4
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c-2=0,,,,c=2
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c-7=0,,,,c=7
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in order,,,,,-4,,,,2,,,,,,7
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now lets check numbers before and after each of these against original inequality
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for x less than (-4),,, let x=-5,,, -84 >0,,,, not true
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for -4 less than x less than 2,,,let x=0,,,,,56>0,,,,,ok
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for 2 less than x less than 7,,,,,let x= 3,,,,,-28>0,,,, not true
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for x>7,,,,,let x = 8,,,,,72>0,,,ok
.
therefore ,,,,-4less than x less than 2,,,,x>7,,,are answers

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4k^3 -6k^2 < = 4k
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lets solve as equality first
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4k^3 -6k^2 = 4k
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4k^3 -6k^2 -4k =0
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2k ( 2k^2 -3k -2) = 0
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2k (2k+1)(k-2) =0
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2k =0,,,,k=0
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2k+1 =0,,,,k= (-1/2)
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k-2 = 0,,,,k=2
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Now on number line we have,,, (-1/2),,,,(0),,,,,(2)
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Now lets check if numbers before and after each of these satisfy original inequality
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for x< (-1/2),,,,let x = -1,,,,-10 <= -4,,,,ok
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for (-1/2) < x < 0,,,let x= (-.25),,,-.3125 <=-1,,,,not true
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for 0 less than x less than 2,,,,,let x=1,,,,-2 <= 4,,,,ok
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for x>2,,,,let x= 3,,,,,54<=12,,,,,not true
.
Therefore,,,,,x<=(-1/2),,,,,0<=x<=2,,,,,answer

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Many different ways to solve this ,,,but lets just match form, y=mx +b
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2x -4y = -12
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-4y = -2x -12
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y = (1/2)x + (3),,,,,or m=1/2,,,b=3
.

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(-6x+2) (3x^2+7) > 0
.
lets solve eqn as if it were equal to zero
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(-6x+2) (3x^2 +7) = 0
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-6x+2 = 0
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-6x = -2
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x= + 1/3,,,,,zero
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3x^2 +7 =0
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3x^2 =-7
x^2 = -7/3
x= +/- 1.53 i,,,,imaginary therefore not a Real Number Solution
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The only Real answer is x= +1/3
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If we had solved this part of the problem with the inequality problem with the inequality sign, -6x+2 >0, the answer would be
x < 1/3.
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Lets check by picking test points on either side of + 1/3
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Smaller than + 1/3, use 0,
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substituting,,,(-6(0)+2)(3(0)^2 +7) = 2*7=14,,,,or 14 > 0 ,,,,ok
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Larger than +1/3 ,,,use 1
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subst,,,,,(-6(1)+2)(3(1)^2 +7) = (-4)(10) = -40,,,,or -40>0,,,,not true
.
Therefore ,,,,,x< 1/3,,,,is valid answer






.

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For this problem, lets first find the slope of the original equation, using y=mx +b form, and then define the slope of the second eqn as, m2=-1/m1,perpendicular. Finally, we can define the eqn of the second line from the slope and the given point.
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-8x +5y =77
5y = 8x +77
y = (8/5)x + 77/5
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or from form ,,,,y= m x +b,,,m = (8/5)
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m2= - 1/m1 = (-5/8)
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Second eqn is , y= mx +b,,,,y = (-5/8) x +b
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As it contains the point ( -4,3),,,,substitute
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(3) = (-5/8) (-4) + b
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3 = 20/8 +b
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24/8 - 20/8 = 4/8 =1/2 = b
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subst m,,and b,,in form eqn,, y = (-5/8)x + (1/2)
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mult thru by 8
.
8y = -5x + 4
.
checking, subst ( -4,3),,,,8*3=(-5)(-4) +4 = 24 ,,,,ok

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Please review to assure correct problem
.
3/(-5-x) - x/(x-3)
.
just rearrange the first term, to get better den
.
-3/(x+5) -x /(x-3)
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LCD = (x+5)(x-3)
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[ -3 (x-3) -x(x+5)] / (x+5)(x-3)
.
[ -3x +9 -x^2 -5x ] / (x+5)(x-3)
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[ -x^2 -8x +9 ] / (x+5)(x-3)
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[ -1 ( x^2 +8x -9) ] / ( x+5)(x-3)
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[-1 (x-1)(x+9)] / [ (x+5)(x-3) ]
.
checking let x=0 in original and answer,,,both = -3/5,,,,ok
doing same for x=1, both = zero,,,ok

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Remember with similar figures, that have proportional sides,
.
length and perimeter varies as Scale Factor ^1
.
Area varies as SF^2
.
and Volume varies as SF^3
.
in our case SF = 4/9, perimeters have this same ratio, 4/9

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3+3/7 + 5 = 2
.
Please review what the problem is ??
.
if it is (3+3) / (7+5) = 6 / (12) = 1/2
.

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f(3) = 7-x^2 = 7 -(3)^2 = 7-9 = -2
.
f(0) = x-3x^2 = (0) -3(0)^2 = 0
.

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Lateral Faces of pyramids are triangles ( not rectangles) therefore false

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12x -7x +10 > 0
5x +10 >0
5x > -10
x> -2

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10x +2 -5x -6 > -2(6-x)
.
10x +2 -5x -6 > -12 +2x
.
5x -4 > -12 +2x
.
3x > -8
.
x> -8/3

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Please check the problem
.
(1) ( 6-3) / (6x) < 8
(3) / (6x) < 8
3 < 48 x
3/48 1/16< x
.
(2) 6 -(3/6)x < 8
-(1/2)x < 2
-x < 4
x > -4,,,note inequality flip
.
.
(3) 6 - {3/(6x) } < 8
-{3/(6x)} < 2
-3 < 12x
-3/(12) < x
- (1/4) < x
.

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(4/9)1/2=4/9*1/2=4/18=2/9

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(-2)^-5 =1/(-2)^5=1/-32 = (-1/32)

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(1) A + B = 3,,,,,,,,B=(3-A)
(2) A^3 + B^3 =36
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A^3 + (3-A)^3 =36
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side problem (3-A)^3 = (3-A) (3-A)(3-A)
(3-A) ( 9-6A +A^2)
27 -27A +9A^2 -A^3
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subst in
.
A^3 +(27 -27A +9A^2 -A^3) =36
9A^2 -27A -9 =0
9(A^2 -3A -1) =0
(A^2 -3A -1) =0
.
To factor use quadratic eqn
.
a=1,,,,,b=-3,,,,,c=-1
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A =[ -(-3) +/- sqrt((-3)^2 - 4 (1)(-1)] / 2(1)
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A= [3 +/- sqrt{ 9+4} ]/2
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A= [3 +/- 3.6]/2
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A= 3.3, -.3,,,,,
.
B= 3-3.3 =-.3,,,,or B= 3-(-.3) = 3.3
.
( A^2 +B^2) - AB =?
.
{ (3.3)^2 +(-.3)^2 } - (3.3)(-.3) = ?
.
11+10 = ?
.
21=?

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Remember the order of operations
.
PEMDAS, Parenthesis, Exponents, Multiply/Divide, lt to rt, Add/Subtract, lt to rt
.
3{ 15 -(8-5) } / (2^2 *3/2),,,,check that we got 3/2 in right spot
.
3{ 15 -3} / (4*3/2)
.
3 { 12) / 6
.
36/6
.
6
.
IF the (3/2) is separate
.
[3{ 15-(8-5)} / 2^2 ] * 3/2
.
[3{15 -3} / 4] *1.5
.
[ 3(12) /4 ] * 1.5
.
(36/4) * 1.5
.
9*1.5
.
13.5
.

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usually we use a "factor tree" to better visualize ( please excuse ,,,,)
.
,,,,,,,,,,,,,,,,,,,,,,,,,120
,,,,,,,,,,,,,,,,,,,,,,,,5,,,,24...........these are factors of 120
,,,,,,,,,,,,,,,,,,,,,,,,,,,,2,,,12........these are factors of 24
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2,,,6......these are factors of 12
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,2,,3....these are factors of 6
.
Therefore prime factors of 120 are 2,2,2,3,5,,(which multiply to 120)
.
better stated as 2^3, 3, 5
.
Note we do not use the 1, in this factorization
.

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2log(b) x -3( log(b) y + log(b) z
.
remember a log c = logc^a
.
log(b)x^2 -log(b)y^3 -log(b)z^3
.
remember log a +log b = log(a*b),,,and log a -log b =log(a/b)
.
log(b) { x^2 / y^3 * z^3 }
.
checking, let b=10,,, a=1, ,, b=2,,,c=3
.
original eqn = -2.33,,,,answer = -2.33,,,,,ok

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Vol = 50x^3 -2x +75x^2 -3
.
looking for V = L * W * H
.
factor original eqn
.
2x(25x^2 -1) +3(25x^2-1)
.
(2x+3)( 25x^2 -1)
.
(2x+3) (5x+1)(5x-1),,,,,sides of box
.
2x+3 =0,,,,,x=-1.5
.
5x +1 = 0,,,,,x= -1/5
.
5x-1=0,,,,,,x= 1/5
.
Smallest zero is (-1.5),, therefore x> -1.5