Step-by-step explanation:
since we are using Pythagoras
c² = a² + b²
with c being the Hypotenuse (the baseline opposite of the 90° angle),
you want to get the expression
y² = 3² + 3²
so, you pick the ² and put it next to the y left of the "=" sign.
and to the right of the "=" sign you put 3, ², +, 3, ²
Answer: 1/7
Step-by-step explanation:
26+6+10=42
6/42=1/7
I and 4i
let's say if you have i you can automatically think there is a one in front of it so one term like this (i) could look like this 1i and any number with the same variable or letter are like terms
hope this helps
Apply the rule: 
![3[2 ln(x-1) - lnx] + ln(x+1)=3[ln(x-1)^{2} - lnx ] + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3%5Bln%28x-1%29%5E%7B2%7D%20-%20lnx%20%5D%20%2B%20ln%28x%2B1%29)
Apply the rule : 
![3[2 ln(x-1) - lnx] + ln(x+1)=3ln\frac{(x-1)^{2} }{x} + ln(x+1)](https://tex.z-dn.net/?f=3%5B2%20ln%28x-1%29%20-%20lnx%5D%20%2B%20ln%28x%2B1%29%3D3ln%5Cfrac%7B%28x-1%29%5E%7B2%7D%20%7D%7Bx%7D%20%2B%20ln%28x%2B1%29)
Apply the rule: 
![3[ln (x-1)^{2} -ln x]+ln (x+1)= ln \frac{(x-1)^{6} }{x^{3} } +log(x+1)](https://tex.z-dn.net/?f=3%5Bln%20%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Bln%20%28x%2B1%29%3D%20ln%20%5Cfrac%7B%28x-1%29%5E%7B6%7D%20%7D%7Bx%5E%7B3%7D%20%7D%20%2Blog%28x%2B1%29)
Finally, apply the rule: log a + log b = log ab
![3[ln(x-1)^{2} -ln x]+log(x+1)=ln\frac{(x-1)^{6}(x+1) }{x^{3} }](https://tex.z-dn.net/?f=3%5Bln%28x-1%29%5E%7B2%7D%20-ln%20x%5D%2Blog%28x%2B1%29%3Dln%5Cfrac%7B%28x-1%29%5E%7B6%7D%28x%2B1%29%20%7D%7Bx%5E%7B3%7D%20%7D)
Answer:
h(2)+g(2) = -3
Step-by-step explanation:

Replace the variable (t) with (
2) in the expression.
h (2) = 3 - 5
Replace the variable (t) with (
2) in the expression.
g(2) = 2(2) -5
Replace the function designators in h(2) +g(2) with the actual functions.
h(t) = 3 - 5 +2 (2) ← plug h(2) into 2(t)
Remove parentheses.
3 - 5 + 2(2)
Multiply 2 by 2
3 - 5 + 4 - 5
Subtract 5 from 3
.
-2 + 4 - 5
Add -2 and 4
2 - 5
Subtract 5 from 2
-3