f.3 math

1. let three consecutive numbers be x, x+1, x+2
and the sum of them is less than 25.
therefore
x+ (x+1) + (x+2) < 25
=> 3x+3 < 25
=> 3x < 22
=> x < 22/3
=> x <= 7

the largest value of these number is 7 , 8, and 9.

2.let the length and width of the rectangle be x, and y respectively
therefore
we hold these equation and inequality,
x = 2y
2x+2y < 48

implied that
2(2y) + 2y < 48
=> 6y < 48
=> y < 8

The upper bound of the area of this rectangle is 128, (= 16*8); however, according to the provided information, the rectangle area can never be 128.

3. Let a and b be the numbers of apple and orange respectively.
2a + 1.5b <= 65
a <= 32
b <= 43

(a) the max. is 43 and the min is 32.
(b) Based on the decision of he/ she made, he/she has spent 17.5 ,(5*(1.5+2)); therefore, the amount to be used become 57.5 ,(65-17.5).
2a+1.5b <= 47.5
a <= 23.75
b <= 31.66
(b) the max. is 41 and the min. is 33, which has already counted the 5 apples and 5 oranges.

***** inequality******
(-8x+1)/2-(x+5)/3<2(x-13(1/4))
multiply both six by 6
therefore
3(-8x + 1) - 2(x+5) < 12(x-13(1/4))
=> -24x + 3 - 2x - 10 < 12x - 53*3
=> -26x - 7 < 12x -159
=> 159 - 7 < 12x + 26x
=> 152 < 39x
=> 152/39 < x

(a) (5)^(-3) = 1/5^3 = 1/125 = 0.008
(b) (1/3)^3 = 1/3^3 = 1/27 = 0.037

the change of percentage.
s2 = (1+20%)s1 = 1.2s1
s3 = (1+20%)s2 = 1.2s2
s3 = 1.2*(1.2s1)
=> s3 = 1.44s1
the change of percentage
(s3 -s1 )/ s1 = (1.44s1 -s1) /s1 = 0.44
the sales of Year 3 comparing to that of Year 1 is increased by 44%.

Amy paid x for the camera.
(a)Amy lose 20%, it means Amy received money 0.8x, (1-20%)x. That is Peter paid for the camera.
(b)Peter gain 10%, it means Peter receiced money 0.88x , 1.1*0.8x, (1+10%)[0.8x]. That is Alice paid for the camera.
(c)0.88x = 2640, and Amy loss is 0.2x, as she sold it on 0.8x.
therefore, amy loss is equal to
0.2*(2640/0.88)
= 600.
Amy loss is $600.