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multiplying by sinx __ (1-cosx) + (sinx^2)/(1-cosx) = 2

substituting __ (1-cosx) + (1-cosx^2)/(1-cosx) = 2

factoring __ (1-cosx) + [(1+cos)(1-cosx)]/(1-cosx) = 2

cancelling __ 1-cosx + 1+cosx = 2 __ 2=2

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vertex form is y=a(x-h)^2+k, where (h,k) is the vertex

y=x^2-16x+56 __ subtracting 56 __ y-56=x^2-16x

completing the square by adding half of the x coefficient, squared __ y-56+(-16/2)^2=x^2-16x+(-16/2)^2

y-56+64=x^2-16x+64 __ y+8=(x-8)^2 __ subtracting 8 __ y=(x-8)^2-8

the vertex is (8,-8)

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you are correct

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letx="1st test", y="2nd test", and z="3rd test"

"Kristen has a total of 225 on three tests" __ x+y+z=225

" The sum of the scores on the first and second tests exceeds her third score by 61" __ x+y=z+61

"her first score exceeds her second by 6" __ x=y+6

substituting from 2nd equation into 1st __ (z+61)+z=225 __ 2z+61=225 __ 2z=164 __ z=82

substituting from 3rd eqn into 2nd __ (y+6)+y=(82)+61 __ 2y+6=143 __ 2y=137 __ y=68.5

substituting into 3rd eqn __ x=(68.5)+6 __ x=74.5

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if you draw a picture, you will see that the height (of the girl or the lamp post)
__ is proportional to the distance from the end of the shadow

let x="height of lamp post"

4/3=x/(3+6) __ 36=3x __ 12=x

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L * W = 63

L + 2W = 25 __ L = 25 - 2W __ substituting __ (25-2W)*W=63 __ 2W^2-25W+63=0

factoring __ (2W-7)(W-9)=0

2W-7=0 __ W=7/2 __ substituting __ L+2(7/2)=25 __ L=18

W-9=0 __ W=9 __ substituting __ L+2(9)=25 __ L=7

the length is more than the width, so the 1st solution is the correct one

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sooo...we're now in calculus.com...

r=h/2 __ V=π(r^2)h/3 __ substituting __ V=π(h^3)/12

differentiating __ dV/dT=(π/12)(3h^2)dh/dT __ dh/dT=(dV/dT)/[(π/12)(3h^2)]

dh/dT=8/(π/4)(6^2)=32/(36π)=.283ft/min (approx)

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the slope is the difference in the y-values divided by the difference in the x-values

in this case __ (160-80)/(12-4)=80/8=10

the slope represents the ratio of the rates of change of y and x

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this being algebra.com...let's use an equation...

let x="kg of walnuts", so 50-x="kg of cashews" __ 50 kg total, right?

4.05(x)+7.25(50-x)=6.25(50) __ 4.05x+362.5-7.25x=312.5 __ -3.2x=-50 __ x=15.6

50-x=34.4

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the sum of the terms for an arithmetic sequence is the number of terms (n) multiplied by the average value of the terms
__ (T1+Tn)/2

S = n(5+5+3(n-1))/2 = n(10+3n-3)/2 = (3n^2+7n)/2

(3n^2+7n)/2 = 25n __ 3n^2+7n=50n __ 3n^2-43n=0 __ n(3n-43)=0

n=0 __ they are together on July 1

3n-43=0 __ 3n=43 __ n=14 1/3 __ John passed Miku on July 15

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the walk is on all four sides of the pool, so the dimensions of the pool are 20-2w and 22-2w
__ where w is the width of the walk

(20-2w)(22-2w)=323 __ 440-84w+4w^2=323 __ 4w^2-84w+117=0

use quadratic formula to find w




answer below


w=1.5

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the general form is y=mx+b, where m is the slope and b is the y-intercept

y=(3/2)x+3

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the easiest way is to think about which unit is smaller or larger
__ with fractional quantities (like ft/sec or dollars/gal), do the numerator and denominator seperately

miles are larger than feet, so there will be fewer for the same distance
__ this means multiplying feet by a fraction LESS than one

the same is true for hours and seconds
__ but in this case, the seconds are in the denominator
__ so a smaller denominator means a larger fraction

44ft/sec __ there are 60 sec in a min, so you can go 60 times as far as in a sec
__ 44ft/sec = 2640ft/min
__ there are 60 min in an hr, so you can go 60 times as far as in a min
__ 2640ft/sec = 158400ft/hr

there are 5280ft in a mi __ 158400ft/hr / 5280ft/mi = 30mi/hr


you can also use "conversion factors" incorporating units
__ the units cancel like variables in multiplying fractions

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subtracting 4-8x __ x^2+8x-4=0 __ this is not factorable

adding 4 __ x^+8x=4 __ completing the square by adding (8/2)^2 __ x^2+8x+16=20

taking square root __ x+4=ħsqrt(20) __ x=-4 ħ 2(sqrt(5))

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possible groups of 3 out of 27 __ 27C3 __ 27*26*25/(3*2*1) __ 2925

possible groups of 2 red out of 14 __ 14C2 __ 14*13/(2*1) __ 91

possible groups of 1 blue out of 13 __ 13

probability of 2 red and 1 blue __ 91*13/2925 __ .404 (approx)

possible groups of 3 red out of 14 __ 14C3 __ 14*13*12/(3*2*1) __ 364

probability of 3 red __ 364/2925 __ .124 (approx)

possible groups of 3 blue out of 13 __ 13C3 __ 13*12*11/(3*2*1) __ 286

probability of 3 blue __ 286/2925 __ .098 (approx)

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possible groups of 3 out of 27 __ 27C3 __ 27*26*25/(3*2*1) __ 2925

possible groups of 2 red out of 14 __ 14C2 __ 14*13/(2*1) __ 91

possible groups of 1 blue out of 13 __ 13

probability of 2 red and 1 blue __ 91*13/2925 __ .404 (approx)

possible groups of 3 red out of 14 __ 14C3 __ 14*13*12/(3*2*1) __ 364

probability of 3 red __ 364/2925 __ .124 (approx)

possible groups of 3 blue out of 13 __ 13C3 __ 13*12*11/(3*2*1) __ 286

probability of 3 blue __ 286/2925 __ .098 (approx)

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possible groups of 3 out of 27 __ 27C3 __ 27*26*25/(3*2*1) __ 2925

possible groups of 2 red out of 14 __ 14C2 __ 14*13/(2*1) __ 91

possible groups of 1 blue out of 13 __ 13

probability of 2 red and 1 blue __ 91*13/2925 __ .404 (approx)

possible groups of 3 red out of 14 __ 14C3 __ 14*13*12/(3*2*1) __ 364

probability of 3 red __ 364/2925 __ .124 (approx)

possible groups of 3 blue out of 13 __ 13C3 __ 13*12*11/(3*2*1) __ 286

probability of 3 blue __ 286/2925 __ .098 (approx)

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let x="distance from wall"

"The top of a 15-foot ladder is 3ft farther up a wall than the foot of the ladder is from the bottom of the wall"
__ x+3

by Pythagoras __ x^2+(x+3)^2=15^2 __ x^2+x^2+6x+9=225 __ 2x^2+6x-216=0 __ x^2+3x-108=0

factoring __ (x+12)(x-9)=0

x+12=0 __ x=-12 __ negative value not realistic

x-9=0 __ x=9

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using rules for logs __ (3x-2)(x-3)=6

3x^2-11x+6=6 __ 3x^2-11x=0 __ x(3x-11)=0

x=0 __ not realistic __ logs are not defined for negative values

3x-11=0 __ x=11/3

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this is the log of a log

using rules for logs __ log2(x)=12^3 __ x=2^(12^3) __ x=2^1728 __ x=1.513x10^520 (approx)

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since 2+3=5, divide PD into 5 parts with PT equal to 2 parts and TD equal to 3 parts

this is very straight forward because the x-distance and the y-distance are both divisible by 5

Tx=Px+(2/5)(7-2) __ Tx=2+2 __ Tx=4

Ty=Py+(2/5)(11-1) __ Ty=1+4 __ Ty=5

T=(4,5)

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the x-intercept is when the graph crosses the x-axis __ at this point, y=0
__ similarly, the value of x is zero at the y-intercept

to find the x-intercept, substitute zero for y; and vice versa

x-intercept __ 12(0)+15(0)+21=4x-5(0)+7 __ 21=4x+7 __ 14=4x __ 3.5=x __ so the x-intercept is (3.5,0)

y-intercept __ 12y+15y+21=4(0)+5y+7 __ 27y+21=5y+7 __ 22y=-14 __ y=-7/11 __ so the y-intercept is (0,-7/11)

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adding the 2nd eqn to the 3rd eqn ELIMINATEs the y __ 3x-3z=6

adding two times the 2nd eqn to the 1st eqn also eliminates the y __ 3x+z=14

subtracting the 2nd NEW eqn from the 1st NEW eqn eliminates the x __ -4z=-8 __ z=2

substituting __ 3x+(2)=14 __ x=4

substituting __ (4)+2y+3(2)=8 __ 2y=-2 __ y=-1

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a general form for the equation of a parabola is __ (x-h)^2=4p(y-k)
__ where (h,k) is the vertex and p is the distance from the vertex to the focus

the vertex is midway between the focus and the directrix
__ in this case, (4,1)
__ so p=-1

the equation would be (x-4)^2=-4(y-1)

solving for y gives y=(-1/4)(x-4)^2 + 1

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to find the population at the end of any year, you take 8% of the starting population and add it

P2=P1+(8%)P1 __ factoring __ P2=P1(1+8%)=P1(1.08)

P3=P2+(8%)P2 __ factoring __ P3=P2(1+8%)=P2(1.08) __ substituting P3=(P1(1.08))(1.08) __ P3=P1(1.08)^2
__ 2 is the number of years between P1 and P3

in general __ Pn=P0(1.08)^n __ where P0 is the original starting population


200=100(1.08)^n __ dividing by 100 __ 2=1.08^n __ taking log __ log(2)=n(log(1.08))

dividing by log(1.08) __ (log(2))/(log(1.08))=n __ 9=n (approx)

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cost times number equals receipts

"total receipts were $824" __ 3a+2s=824

"twice as many adult tickets as student tickets were sold" __ a=2s

substituting __ 3a+(a)=824 __ 4a=824 __ a=206

substituting __ 206=2s __ 103=s

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eliminate the same variable from two different PAIRS of equations, leaving two new equations with two unknowns

multiply 2nd eqn by -3 and add to 1st eqn __ -7y+10z=-21

multiply 3rd eqn by -3 and add to 2nd eqn __ 11y-9z=33

multiplying 1st new eqn by 11 __ -77y+110z=-231

multiplying 2n new eqn by 7 __ 77y-63z=231

adding eqns __ 47z=0 __ z=0

substituting __ 11y-9(0)=33 __ y=3

substituting __ x-3(3)+2(0)=-12 __ x-9=-12 __ x=-3

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d=rt __ distance out is the same as distance back

the total time is 7hr __ if t is the time out, then 7-t is the time back

rate with the wind is 280+40 or 320

rate back is 280-40 or 240

320 * t = 240 * (7-t) __ 320t=1680-240t __ 560t=1680 __ t=3

substituting __ d=320(3) __ d=960 (one way)

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"A square and a rectangle have the same perimeter" __ 4S=2L+2W

"The lenghth of the rectangle is 4cm less than twice the side of the square" __ L=2S-4

"the width of the rectangle is 6cm less than the side of the square" __ W=S-6

substituting __ 4S=2(2S-4)+2(S-6) __ 4S=4S-8+2S-12 __ 4S=6S-20 __ 20=2S __ 10=S

so the perimeter is 4*10 or 40

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distributing the 3 __ 3x+6=7x-10

adding 10 __ 3x+16=7x

subtracting 3x __ 16=4x

dividing by 4 __ 4=x

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you are looking for the lowest common multiple

45sec = 3 * 15

2min = 120sec = 8 * 15

1 1/4 min = 75sec = 5 * 15

LCM = 15 * 3 * 8 * 5 = 1800sec = 30min

so, every half hour they flash together