x2 + 14x + 45 = (x + 9) (x + 5)
3 with 3 remaining45 - 3 = 42 = 14 x 3
14 out of 45 = 14/45 = 0.3111...
in algebra terms, he is x-7 now, so 7 years ago is was x-14
X-45=12 ________ +45 +45 ________ x= 57
-x2-14x-45 = -(x2+14x+45) delta = 142-4*1*45=16 so x2+14x+45=(x-(-14+4)/2) (x-(-14-4)/2) = (x+5)(x+9) so -x2-14x-45 = -(x+5)(x+9)
5 & 9. 5 + 9 = 14... 5 x 9 = 45.
x2 + 14x + 45 = (x + 9) (x + 5)
Let's denote the two numbers as x and y. We know that xy = 45 and x + y = 14. To find the two numbers, we can set up a system of equations. From the first equation, we can express y in terms of x as y = 45/x. Substituting this into the second equation gives us x + 45/x = 14. Multiplying through by x gives us x^2 - 14x + 45 = 0. Factoring this quadratic equation gives us (x - 9)(x - 5) = 0. Therefore, the two numbers are x = 9 and y = 5.
Assuming the numbers are organized as listed below, here is the answer: (98 / 7) x (45 + 35) = 14 x 80 = 1,120 Alternate Possibility: [(98 / 7) x 45] + 35 = (45 x 45) + 35 = 2025 + 35 = 2,060
3 with 3 remaining45 - 3 = 42 = 14 x 3
14 out of 45 = 14/45 = 0.3111...
59
1 x 45 = 45 2 x 45 = 90 3 x 45 = 135 4 x 45 = 180 5 x 45 = 225 6 x 45 = 270 7 x 45 = 315 8 x 45 = 360 9 x 45 = 405 10 x 45 = 450 11 x 45 = 495 12 x 45 = 540
in algebra terms, he is x-7 now, so 7 years ago is was x-14
The first ten positive integer multiples of 45 are as follows: 1 x 45 = 45 2 x 45 = 90 3 x 45 = 135 4 x 45 = 180 5 x 45 = 225 6 x 45 = 270 7 x 45 = 315 8 x 45 = 360 9 x 45 = 405 10 x 45 = 450
X-45=12 ________ +45 +45 ________ x= 57