Question 147167
1---It is a square.  The diagonals of a square are just the length of any side multiplied by {{{sqrt(2)}}}.  This is true because the diagonal bisects the square into two 45, 45, 90 degree triangles, (since it also bisects two angles).  The hypotenuse of a 45,45,90 triangle is always a side times radical 2. The hypotenuse of the triangle is the diagonal of the square.
2---Find the length from one side to the midpoint (if you graph this, it will be easier).  Then add that length onto it in your graph.  See what point you land on and record it in (x,y) form.
3---Solution for t is anything zero and up in this case. Since the expression that contains it has to be more then one and the denominator cannot be zero.  Interval notation would just be the increases and decreases of the graph.  Basically in which values of x does y decrease, and which values of x does y increase. The formula for the notation is D or I:(lowest x value,highest x value).  Most likely in your interval notation a min or max point will be infinity.
4---{{{(4x+3)-11<1/2}}}={{{x-2<1/8}}}.  All you have to do to solve this is combine like terms in the first equation and get {{{4x-8<1/2}}}.  You can simplify this by taking 4 out of all the terms in the expression. Divide everything by 4 and get {{{x-2<1/8}}} the same as the other expression.