I am unsure about the very last problem but I can help with the first two
1) (y+1)+4
If we combine the numbers 1 and 4, we get +5 and can isolate the numbers from the variable.
This would give us

2) (6*r)*7
remember that we do not have to explicitly state 6*r
Instead, we can write it as 6r
this helps us get rid of the parentheses
now we can write it as

I hope this helps!:)
Since a pentagon has 5 sides, you would divide the perimeter by the number of sides.
641.5/5 = 128.3cm
Therefore, the length of one side of the pentagon is 128.3cm.
Wouldn’t that just be 325m=c
Then for the second part it’s just 325(7)=c
Multiply 325 times 7 and that’s the answer
-2/3x+9=4/3x-3 original problem add 2/3x to both sides
9=6/3x-3 simplify 6/3 to 2 and add 3 to both sides
12=2x divide both sides by 2
x=6
hope that helps
Begin by finding the lowest point the quadratic equation can be, the vertex;
x²-1= is just a translation down of the graph x²
vertex; (0, -1) and since the graph of x² would extend to infinity beyond that point, we can say {x| x≥0} for domain and {y| y≥-1}.
For the linear equation, it is possible to have all x and y values, therefore range and domain belong to all real numbers.
Hope I helped :)