2+½, is just 2½, now divided by 1/4.
let's first convert the mixed fraction to improper, and then divide.
![\bf \stackrel{mixed}{2\frac{1}{2}}\implies \cfrac{2\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{5}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{5}{2}\div \cfrac{1}{4}\implies \cfrac{5}{2}\cdot \cfrac{4}{1}\implies \cfrac{5}{1}\cdot \cfrac{4}{2}\implies 5\cdot 2\implies 10](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B2%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B2%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B5%7D%7B2%7D%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill%5C%5C%5C%5C%0A%5Ccfrac%7B5%7D%7B2%7D%5Cdiv%20%5Ccfrac%7B1%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B2%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B1%7D%5Cimplies%20%5Ccfrac%7B5%7D%7B1%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B2%7D%5Cimplies%205%5Ccdot%202%5Cimplies%2010)
Answer:
it's the first one girl! :)
Step-by-step explanation:
I wrote my decision on a piece of paper.
Answer:
(3, 2), (2, 3)
Step-by-step explanation:
x + y = 5
xy = 6
Solve the first equation for x.
x = 5 - y
Substitute 5 - y for x in the second equation.
xy = 6
(5 - y)y = 6
5y - y² = 6
y² - 5y + 6 = 0
Factor.
(y - 2)(y - 3) = 0
y - 2 = 0 or y - 3 = 0
y = 2 or y = 3
Now substitute 2 for y in the first equation and solve for x.
x + y = 5
x + 2 = 5
x = 3
One solution is x = 3; y = 2, or (3, 2).
Now substitute 3 for y in the first equation and solve for x.
x + y = 5
x + 3 = 5
x = 2
Another solution is x = 2; y = 3, or (3, 2).
Answer: (3, 2), (2, 3)
You haven't shared the possible answers, so the best I can do (which is very good!) is to assume we want to change from base 4 to base 10 and then apply the change of base formula.
Given log-to-the-base-4-of (x+2), we want log-to-the-base-10 of (x+2). Following the change of base formula,
log-to-the-base-4-of (x+2)
log-to-the-base-10 of (x+2) = ------------------------------------
log-to-the-base-4-of-10
