let's firstly convert the mixed fractions to improper fractions and then to do away with the denominators, let's multiply both sides by the LCD of all denominators.
![\stackrel{mixed}{1\frac{3}{4}}\implies \cfrac{1\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{7}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{7}{4}-\cfrac{4}{5}=\cfrac{35}{20}-\boxed{?}\implies \stackrel{\textit{multipling both sides by }\stackrel{LCD}{20}}{20\left( \cfrac{7}{4}-\cfrac{4}{5} \right)=20\left( \cfrac{35}{20}-\boxed{?} \right)} \\\\\\ 35-16=35-20\boxed{?}\implies 19=35-20\boxed{?}\implies -16=-20\boxed{?} \\\\\\ \cfrac{-16}{-20}=\boxed{?}\implies \cfrac{4}{5}=\boxed{?}](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B1%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B1%5Ccdot%204%2B3%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B7%7D%7B4%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B7%7D%7B4%7D-%5Ccfrac%7B4%7D%7B5%7D%3D%5Ccfrac%7B35%7D%7B20%7D-%5Cboxed%7B%3F%7D%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bmultipling%20both%20sides%20by%20%7D%5Cstackrel%7BLCD%7D%7B20%7D%7D%7B20%5Cleft%28%20%5Ccfrac%7B7%7D%7B4%7D-%5Ccfrac%7B4%7D%7B5%7D%20%5Cright%29%3D20%5Cleft%28%20%5Ccfrac%7B35%7D%7B20%7D-%5Cboxed%7B%3F%7D%20%5Cright%29%7D%20%5C%5C%5C%5C%5C%5C%2035-16%3D35-20%5Cboxed%7B%3F%7D%5Cimplies%2019%3D35-20%5Cboxed%7B%3F%7D%5Cimplies%20-16%3D-20%5Cboxed%7B%3F%7D%20%5C%5C%5C%5C%5C%5C%20%5Ccfrac%7B-16%7D%7B-20%7D%3D%5Cboxed%7B%3F%7D%5Cimplies%20%5Ccfrac%7B4%7D%7B5%7D%3D%5Cboxed%7B%3F%7D)
Answer:
See answers below
Step-by-step explanation:
T59 = a+58d = -61
T4 = a+3d = 64.
Subtract
58d-3d = -61-64
-55d = -125
d =125/55
d = 25/11
Get a;
From 2
a+3d = 64
a+3(25/11) = 64
a = 64-75/11
a = 704-75/11
a = 629/11
T23 = a+22d
T23 = 629/11+22(25/11)
T23 = 1179/11
The answers would most likely be A,C,and B from my calculations
So, this question is basically asking us "If we had an x instead of a 2, would this be true?" We can try and see what we get:

So, if we want to show this we have to change the numerator or denominator in such a way that we can cancel some common factors. Notice that 
If we replace the factored numerator with the original one, we get:

Since we have an equality, this relation is proved.
Answer:
0.1875
Step-by-step explanation:
Hope it helps you :>
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