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not only possible, but guaranteed ... anything equals itself

the point here is that x does not have a unique value ... any value will satisfy the equation

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c is the supplement of 60° __ c=120°

g is vertical to a 60° angle __ g=60°

f+40°+60°=180° __ f=80°

d+110°+50°=180° __ d=30°

e+60°+80°+30°=180° __ e=10°

a=h __ a+h+40°=180° __ a=h=70°

b=60°+70° __ b=130°

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each ray makes two angles from one angle, so each additional ray means one additional angle

11 rays means 11 angles in addition to the original angle, so 12 angles

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common factors first __ 8(x^2-25y^2)

difference of squares __ 8(x+5y)(x-5y)

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common factors first __ 7(x^2+7x+6)

factors of 6 that "combine" to give 7 __ 7(x+6)(x+1)

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common factors first __ 2a(8a^2-18ay+7y^2)

factors of 8 and 7 that "combine" to give -18 __ 2a(2a-y)(4a-7y)

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common factors first __ 2a(8a^2-18ay+7y^2)

factors of 8 and 7 that "combine" to give -18 __ 2a(2a-y)(4a-7y)

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y=mx+b

y=6x-3

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common factors first __ x^8*y(x^2*y^2-4xy-21)

x^8*y(xy-7)(xy+3)

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(x+8)(x+1)

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common factors first __ 8(x^2-25y^2)

difference of squares __ 8(x+5y)(x-5y)

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(7x+2)(x+8)

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common factors first __ 7(x^2+7x+6)

7(x+6)(x+1)

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you are trying to maximize z

test the vertices by substituting into z=2x+4y to find the maximum

some points can be eliminated by inspection __ (0,0) is obviously not the maximum
__ (6,3) must be greater than each of the first five points in the list
__ the maximum appears to be one of the last three points in the list

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the slope (m) is -1/3 (not 1/3)

substituting to find the y-intercept (b) __ -3=(-1/3)(-5)+b __ -3=(5/3)+b __ -14/3=b

y=(-1/3)x-(14/3)

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let x="A time alone" and y="B time alone"

"Workman A and B complete a certain job if they work together for 6 days"
__ 6/x+6/y=1

"A alone works for 3 days and B alone works for 10 days" __ 3/x+10/y=1
__ multiplying by 2 __ 6/x+20/y=2

subtracting equations __ 6/x+20/y-(6/x+6/y)=2-1 __ 14/y=1 __ 14=y

substituting __ 6/x+6/14=1 __ subtracting 6/14 __ 6/x=8/14 __ multiplying by 14x __ 84=8x
__ dividing by 8 __ 10.5=x

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depending on how you look at the problem...

if k is a digit, then 52^2=2704


if the perfect square is the product of 270 and k

prime factors of 270 __ 3*3*3*5*2
__ for a perfect square, even quantities of all prime factors

3*3*3*5*2*(3*5*2) __ 270*30=8100=90^2

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two children cross __ one rows back __ one adult crosses __ child rows back

repeat cycle until all adults are across (7 more times)

last trip is two children crossing to join 8 adults

4*8+1 or 33 crossings

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call the bales (in order of increasing weight) A, B, C, D, E

the lowest and highest weights are the 2 lightest and 2 heaviest respectively
__ A+B=110 __ D+E=121

the 2nd lowest is the lightest bale with the middle bale __ A+C=112

the 2nd highest is the heaviest bale with the middle bale __ C+E=120

every bale is weighed with each of the other 4 bales, so every bale is weighed 4 times
__ this means that the sum of the weights is 4 times the sum of the weights of the individual bales

A+B+C+D+E=(110+112+113+114+115+116+117+118+120+121)/4 __ A+B+C+D+E=289

substituting __ 110+C+121=289 __ C=58

substituting __ A+58=112 __ A=54

substituting __ 54+B=110 __ B=56

substituting __ 58+E=120 __ E=62

substituting __ D+62=121 __ D=59

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let x=time __ 2[(6-x)/6]=(9-x)/9 __ multiplying by 9 __ 3(6-x)=9-x __ 18-3x=9-x

adding 3x-9 __ 9=2x __ 4.5=x

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using rules of logs (and exponents) __ log(a^2)=2log(a) __ so log(a^2)=.5

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log3(9^(1/5)) __ log3[(3^2)^(1/5)] __ log3(3^(2/5))=2/5

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to eliminate one of the variables, just arrange for the number value of the coefficients of the variable to be equal
__ then add or subtract the equations as needed

1) the number value (absolute value) of the y-coefficients is the same (1)
__ adding the two equations ELIMINATES the y giving 5x+3x=12+4 __ 8x=16 __ x=2
__ substituting __ 3(2)+y=4 __ 6+y=4 __ y=-2

2) the value of the x-coefficients is the same (2)
__ subtracting the 2nd equation from the 1st gives -y-(-2y)=-6-(-4) __ y=-2
__ substituting __ 2x-(-2)=-6 __ 2x=-8 __ x=-4


you can "arrange" for the coefficients to be equal by multiplying one or both of the equations by the appropriate quantities

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yes __ good work

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close...

whether you select red-blue or blue-red, the result is the same

so half of 20 is the answer

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let t="time in minutes" __ 10^8=1[20^(t/5)] __ taking log __ log(10^8)=(t/5)log(20)

dividing by log(20) __ 8/log(20)=t/5 __ multiplying by 5 __ 40/log(20)=t

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1st one yes, 2nd one no

2nd should be (-2,3)

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the number is right but the sign is wrong __ -9 is the y-coordinate

a general equation is __ where (h,k) is the center

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let x="amount @ 3%", y="amount @ 4%", and z="amount @ 6%"

"A man divides $10000 among 3 investments" __ x+y+z=10000

"His annual income from the first two investments is $80 less than his income from the third investment"
__ (3%)x+(4%)y=(6%)z-80

"his total income is $460 per year" __ (3%)x+(4%)y+(6%)z=460

substituting __ [(6%)z-80]+(6%)z=460 __ adding 80 __ (12%)z=540 __ dividing by 12% __ z=4500

substituting __ x+y+(4500)=10000 __ subtracting 4500 __ x+y=5500 __ subtracting x __ y=5500-x

substituting __ (3%)x+(4%)y=(6%)(4500)-80 __ (3%)x+(4%)y=190 __ dividing by 4% __ .75x+y=4750

substituting __ .75x+(5500-x)=4750 __ subrtacting 5500 __ -.25x=-750 __ dividing by -.25 __ x=3000

substituting __ (3000)+y=5500 __ subtracting 3000 __ y=2500

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if n is a root (solution), then x-n is a factor

(x-3)(x+5)=0 __ FOILing __ x^2+2x-15=0

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let 3x="number of animals"

x/2+x/3+x/8=69 __ multiplying by 24 (LCD) __ 12x+8x+3x=69*24 __ 23x=69*24

dividing by 23 __ x=72 __ substituting __ 3(72)=216