I think associative, because it is the same thing in a different order.
Answer:
![[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]](https://tex.z-dn.net/?f=%5Bp-%7Cp%7C%2A10%5E%7B-3%7D%20%5C%2C%20%2C%20%5C%2C%20p%2B%7Cp%7C%2A%2010%5E-3%5D)
Step-by-step explanation
The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is
r = |p*-p|/|p|
we want r to be at most 10⁻³, thus
|p*-p|/|p| ≤ 10⁻³
|p*-p| ≤ |p|* 10⁻³
therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]
Area is length times width. Therefore, width is area divided by length. Convert 11 7/8 to the improper fraction 95/8 and convert 4 3/4 to 19/4. Then, you can fine 95/8 divided by 19/4 by flipping the second fraction and multiplying so that you have 95/8 times 4/19. 4 times 95 is 380 and 8 times 19 is 152. Therefore, the answer is 380/152. This is equivalent to 5/2 or 2.5.
Perpendicular line has slope that multiplies to -1
paralel line has same slope
y=mx+b
slope=-7/4
perpendicular
4/7
(-7,-2)
-2=4/7(-7)+b
-2=-4+b
add 4 both sides
2=b
y=4/7x+2
paralel
y=-7/4x+b
-2=-7/4(-7)+b
-2=49/4+b
minus 49/4 from both sides (-2=-8/4)
-57/4=b
y=-7/4x-14.25
perpendicular
y=4/7x+2
paralel
y=-7/4x-14.25
1. in picture 1
2. in picture 2
Hope this helps :T
Also I am not so shure about 2, but I am 98% shure 1 is right.