Well remember your order of operation (9 * 2) + (25 * 2) = 18 + 50 = 68 hope this helps.
9
9xx-30x+25sqrt 9 = 3, -3sqrt 25 = 5, -52(3*5)=302(-3*-5)=302(-3*5)=-302(3*-5)=-30(5-3x)(5-3x)(3x-5)(3x-5)YES it is a perfect square
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81
25x2+30x+9 25*9=225 15*15=225 15+15=3025x2+15x+15x+9[5*5xx 3*5x] [3*5x 3*3]5x(5x+3)+3(5x+3)(5x+3)(5x+3)Answer: (5x+3)2
Well remember your order of operation (9 * 2) + (25 * 2) = 18 + 50 = 68 hope this helps.
Given ef is the midsegment of isosceles trapezoid abcd bc equals 17x ef equals 22.5x plus 9 and ad equals 30x plus 12 find ad?
9
5x + 9y = 06x + 6y = 2 use substitution method5x + 9y = 0y = -5x/96x + 6y = 2 substitute y with -5x/96x + 6(-5x/9) = 26x - 30x/9 = 2 multiply by 9 to both sides54x - 30x = 1824x = 18x = 24/18 = 4/3y = -5x/9 = (-5/9)x substitute x with 4/3y = (-5/9)(4/3) = -20/27Thus the solution is x = 4/3 and y = -20/27.
1st circle: x^2 +y^2 -6x +8y -75 = 01st circle: center at (3, -4) and radius 102nd circle: x^2 +y^2 -30x -24y +269 = 02nd circle: center at (15, 12) and radius 10Midpoint of (15, 12) and (3, -4): (9, 4) which is the point of contactProof: distance from (15, 12) to (9, 4) = 10Proof: distance from (3, -4) to (9, 4) = 10
9xx-30x+25sqrt 9 = 3, -3sqrt 25 = 5, -52(3*5)=302(-3*-5)=302(-3*5)=-302(3*-5)=-30(5-3x)(5-3x)(3x-5)(3x-5)YES it is a perfect square
That is 27 x 9 = 243
189
The sum of 27 occurrences of the number 9 is calculated by multiplying 27 by 9, resulting in 243.
263
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