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n= p(1)^a(1) + p(2)^a(2) + p(3)^a(3) + .........+ p(k)^a(k)
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n is a natural number,,,this leads to n=1 or larger integer. ,,,, AND it limits each term of the sequence to a positive number. That is, the p"s cannot be + or -, as this would restrict the terms to an even number, to make n positive, which was not stated.
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Likewise, the "a" values cannot be negative, leading to a fraction, or a fraction, which would lead to not whole numbers.
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The given that p(k) is a distinct prime, leads to the min value of p(k) being the min prime, or 2 .,,,or larger,,, (1 is a special number, not prime)
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n = something positive * a number 2 or larger raised to a positive number( making it >2)
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n>2^ k
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and taking log of both sides
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log n > log 2^k or log n > k log 2

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Probability is the ratio of the number going the desired way, to the total possible numbers'
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something that will not happen has 0 chances / total possible chances., or P=0
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If something is certain, we know all of the chances will be in numerator / same number = 1
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And the P = 1,

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Let v = # VIP Rings ,,,,and s = # SST Rings
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(1) 3 v + 2 s = 60 hrs
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(2) v + s = 24,,,,,or v = 24 -s
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subst in (1),,,, 3(24-s) +2 s = 60
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72 -3s +2s =60
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12 = s
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v = 24-s=24-12=12,,,,which is answer (C)
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Profit from 12 each is 12(40) + 12 (35) = $ 900, made within guidelines
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Checking the other options
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A) 14(40) +10(35) =$ 910 BUT
checking hrs, 14(3) + 10(2) = 62 hrs,,,,over limit of 60
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B) 12(40) + 8(35) = $ 760,,,,, low
with hrs of 12(3) +8(2) = 52,,, also low

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A(x) = x^2 +5x +6
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a(5) = (5)^2 + 5(5) +6 = 56
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A(x) = x^2 +5x +6 = 1/2 ( h) (l) = (1/2) (x+3) h,,,,,,,area of triangle = 1/2 base * height
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{ (x+3)(x+2) } / (1/2) ( x+3) = 2(x+2) = h
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check
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A(5) = 56, (x+3)= 8,, A = 1/2 * 8 * h = 56,,,,,h= 112/8 = 14
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h=2(x+2) = 2 (5+2) =14,,,,,,ok

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3x -2y =0
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to find y intercept ,,,let x =0 ( and vice versa )
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at x=0, y=0,,,y intercept is y=0

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when there are four terms, try factoring by grouping
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2x^3 -14x +3x -21
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2x (x-7) +3(x-7},,,,,,like ab +cb= (a+c)b ,,,but b=(x-7)
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(2x+3) (x-7)
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check by subst x=1 in original and answer, ok

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for 3 times around, let Distance = 3 * Circumference
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d = 3 * 2 *pi * radius, radius = 1500 / 2 = 750
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d = 3 * 2 * 3.14 * 750 = 14,130 ft

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If total budget is $ 520,000, and taxes are 15 %
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.15 * $520, 000 = $ 78,000
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check,,,,78 /520 =.15 ,,,,ok
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to factor ,,,x^3 -4x^2 +x +6,,,,knowing (x+1) is a factor,,,,use poly division or synthetic division
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,,,,,,-1,,,,1,,,,-4,,,,,+1,,,,6
,,,,,,,,,,,,,,,,,,,-1,,,,,+5,,,-6
,,,,,,,,,,,,,,,__________________
,,,,,,,,,,,,,,1,,,,-5,,,,+6,,,,,0
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(x+1)(x^2 -5x +6),,,,,,,now just factor 2nd term
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(x+1) ( x-3) (x-2),,,,,or answer B
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to check, multiply factors to reach original,,,,ok

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difference is subtraction
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( n ) - ( k^2 )
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therefore c) n-k^2
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Please see answer to problem # 199732 thanks

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Please see answer to problem # 199732 thanks

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Please see answer to problem # 199732 thanks

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Please see answer to problem # 199732 thanks

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Please see answer to problem # 199732 thanks

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please see solution to problem # 199732 thanks

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1. given
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2 definition of congruence (sides of equal length)
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3 definition of an isosceles triangle ( 2 sides equal)

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(1) 4x +5x -10 = 35
9x -10 =35,,,,,combining like terms
9x =45,,,,,,,,adding 10 to both sides
x=5,,,,,,,,,,,,,dividing by 9
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(2) 9x+6x-12x = -7x +2x -12 +5x
3x = -12,,,,,,,,,combining like terms
x= -4,,,,,,,,dividing by 3
.

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Very interesting question
Let's start with a rough sketch of a circle, now one chord, now another perpendicular chord that goes thru center.
This is the situation that answers your question, yes it can happen but it does not have to.
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The principle that puts the second chord thru the center is,,,, it must be a perpendicular bisector. ALL perp bisectors go thru center, hence could be diameters.
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It is easy to take your sketch, and just slide the second chord to one side or the other. It does not go thru center, and is not a diameter. Your problem did not specify bisection.
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Another interesting construction, again draw a rough circle and one chord. Draw a second chord, perpendicular to first but located at an end of the first chord. NOW, if we connect the other ends of the chord, we have a triangle. But, a very special triangle, it is a right inscribed triangle, that divides the circle into two semicircles, and has the hypotenuse as the diameter.
Good luck with class

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[ 7^(n+2) -35*7^9n-1)] / [ 44(7^(n+2) ) ] = 1/49
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[ 1-35*7^-3] / 44 = 1/49
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[ 1 - 35 / (49*7) ] / 44 = 1/49
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[ 1-5/49] / 44 = 1/49
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[49/49 -5/49 ] / 44 = 1/49
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[ 44 / 49 ] / 44 = 1/49
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(44/49) * (1/44) = 1/49 = 1/49,,,,,,ok

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Remember negative exponent is reciprocal,,,,a^-2 = 1/a^2
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AND , to divide common bases,,,substract exponents,,,,a^4/a^2=a^(4-2) = a^2
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(1) (r^-4)(-6r^2t^2) = -6 r^-2t^2= -6t^2 / r^2
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(2) w^3 / (-6w^-6) = w^9/-6 = (-1/6) w^9
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(3) k^6 / (k^3)^5 = k^6/k^15 =k^-9 = 1/k^9
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(4) t^-9 / q^-9 = q^9/t^9 = (q/t)^9
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check by assigning numbers to variables and checking original and answer,,,all ok

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4(p-9) =64
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p-9 = 64/4 = 16
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p =16 +9 =25
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check
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4(25-9) =64
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4(16) =64,,,,,ok

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please see problem 199342

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y * (1-x) = z
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y-xy =z
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-xy = z -y
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x = (z-y) / ( -y) = -z/y +1 =1-z/y
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check by letting x=4, y=2, then z=-6,,,,ok for both original and answer

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-6r +s -rs
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r(-6-s) +s
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5(-6-7) +7
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5(-13) +7
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-65 +7
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-58
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check using direct subst into initial eqn
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-6(5)+(7) -(5)(7) = -30 +7 -35 = -65 +7 =-58,,,,ok

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Remember,,, form y= a(n)x^n +..... / b(m) x^m +......
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for VERTICAL ASYMPTOTES,,,,,,find zero"s of denominator
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in all 3 cases above,,, (x-2) =0,,,or x=2 is a vertical asymptote
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For HORIZONTAL ASYMPTOTES,,,,
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if n less than m,,,,,,y=0
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if n=m,,,,,y = a(n)/b(m)
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n>m,,,,,,, no horizontal asymptote
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In the 1st case,,, n=0,m=1,,,or n is less than m,,,,,y=0 is hor asymptote
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in the 2nd case,,,n=1,m=1,,or n=m,,,,y = a(n)/b(m) = 1/1 =1,,,, y=1 is hor asymptote
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in the 3rd case, n=2,m=1 ,,or n>m,,,,,therefore no hor asymptote
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for OBLIQUE ASYMPTOTES (SLANT)
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if n=(m+1),,,divide numerator by denominator ,,,,and set answer (without remainder) =y
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in the 3rd case above ,,,n=2,m=1,,or n=(m+1),,,,therefore there is a SLANT asymptote
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( x^2 +x -2) / (x-2) = x+3 +4/(x-2)
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setting y = x+3,,,,,,for slant asymptote
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.We might also look for x intercepts,,,,which are zero's of numerator.
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case 2 ,,,has one at x=-2
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case 3,,,has one at x=-2,,and another at x=+1
.

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There are actually two parabolas that meet this criteria, a "y" parabola, and an "x" parabola
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the standard form for a "y" parabola is ,,,, ( y-k) = a (x-h)^2, where (h,k) is the vertex
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substituting our data,,,, vertex at (5, -7) and (4,1),,,,, ( y-(-7) ) = a (x-5)^2
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@ ( 4,1),,,, (1) +7 =a ( 4-5)^2,,,,8=a(1),,,, a=8
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"y" parabola is y+7 = 8(x-5)^2,,,or y= 8(x-5)^2 -7
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to check substitute in both given points,,,ok
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The standard form for an "x" parabola is,,,,,(x-h) =a ( y-k)^2
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or (x-5) = a (y+7)^2,,,,,and @ (4,1),,,,, (4-5) = a (1+7)^2,,,,a= -1/64
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or (x-5) = -1/64 (y+7)^2,,,,,or x= -1/64 (y+7) ^2 +5
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to check substitute in both sets of points,,,,,,ok

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Let x = velocity in still water
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Distance = Velocity * Time
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D1 = (x+6) 5 = 5x +30
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D2 = (x-6) 10 = 10x -60
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but D1 = D2
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5x +30 = 10x -60
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90 = 5x
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18 = x
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checking
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(18 +6) * 5 = (18-6) 10
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120 =120,,,,ok

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Remember that the volume of a rectangular solid, or prism ,,,,Vol = B * h,,,, where B is the Area of the base,,,,and h is the height
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vol = ( 1.2 *1.2 ) * 2 = 2.88 sq meters
,

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Remember order of operations
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8x +2 = 7(x-4) +6
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8x +2 = 7x -28 +6
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8x -7x = -2 -28 +6
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x= -24
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check
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8( -24) +2 = 7( ( -24) -4 ) +6
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-192 +2 = -196 +6
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-190 = -190,,,,,,ok
.

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Remember that Area = pi * rad^2
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we are given diameter = 42, or radius = 21 ,,,,(42/2)
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Area = pi * r^2 = pi * 21^2 = pi * 441 = 1385.44 inches ^2
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an obscure Area formula is A = pi d^2 /4 = pi 42^2 /4 = 1385.44 also
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