Answer:
C
Step-by-step explanation:
y-3=5(x-2) (rearrange this to be in slope- intercept from) (add 3 to both sides)
y = 5(x-2) + 3 (distribute parentheses)
y = x(5) - 2(5) + 3
y = 5x -10 + 3
y = 5x - 7
recall that for a line with gradient m, the gradient of the perpendicular line will be - (1/m)
hence in our case, our gradient of the original line is 5, hence the gradient of the perpendicular line is -1/5
From the choices, the only one that is consistent with this is C
i.e choice C:
5y + x = 25
5y = -x + 25
y = -(1/5) x + 5 ===> gradient of -1/5
Horizontal translation: 2 units right.
Vertical translation: 5 units up
Stretch/compression: stretched by a factor of 3.
Reflection: not reflected.
Answer:
um 6 times 1.90 = 11.4 hope this helpssssssssssss piece
Step-by-step explanation:
let's firstly convert the mixed fractions to improper fractions, and then subtract.
![\bf \stackrel{mixed}{10\frac{1}{3}}\implies \cfrac{10\cdot 3+1}{3}\implies \stackrel{improper}{\cfrac{31}{3}}~\hfill \stackrel{mixed}{13\frac{1}{2}}\implies \cfrac{13\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{27}{2}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{27}{2}-\cfrac{31}{3}\implies \stackrel{\textit{using the LCD of 6}}{\cfrac{(3)27~~-~~(2)31}{6}}\implies \cfrac{81~~-~~62}{6}\implies \cfrac{19}{6}\implies 3\frac{1}{6}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B10%5Cfrac%7B1%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B10%5Ccdot%203%2B1%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B31%7D%7B3%7D%7D~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B13%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B13%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B27%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Ccfrac%7B27%7D%7B2%7D-%5Ccfrac%7B31%7D%7B3%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20LCD%20of%206%7D%7D%7B%5Ccfrac%7B%283%2927~~-~~%282%2931%7D%7B6%7D%7D%5Cimplies%20%5Ccfrac%7B81~~-~~62%7D%7B6%7D%5Cimplies%20%5Ccfrac%7B19%7D%7B6%7D%5Cimplies%203%5Cfrac%7B1%7D%7B6%7D)
