Notice that

so the constraint is a set of two lines,

and only the first line passes through the first quadrant.
The distance between any point
in the plane is
, but we know that
and
share the same critical points, so we need only worry about minimizing
. The Lagrangian for this problem is then

with partial derivatives (set equal to 0)



We have

which tells us that

so that
is a critical point. The Hessian for the target function
is

which is positive definite for all
, so the critical point is the site of a minimum. The minimum distance itself (which we don't seem to care about for this problem, but we might as well state it) is
.
In this question, we're assuming Ralph had 0 in his saving funds to start with.
The glasses cost 1200 dollars.
There are 12 months in a year.
Each month, Ralph is given his check two times, that means that Ralph receives his check 2(12), or 24, times per year.
Divide 1200 (the cost of the glasses) by 24 (the amount of times Ralph is given his check)
1200 / 24
Divide both numerator and denominator by 12
100 / 2
Simplify
50
Ralph should have put 50 dollars per pay check to pay for the glasses.
A is your answer.
Have an awesome day! :)
This is something you'll need a T table for, or a calculator that can compute critical T values. Either way, we have n = 10 as our sample size, so df = n-1 = 10-1 = 9 is the degrees of freedom.
If you use a table, look at the row that starts with df = 9. Then look at the column that is labeled "95% confidence"
I show an example below of what I mean.
In that diagram, the row and column mentioned intersect at 2.262 (which is approximate). This value then rounds to 2.26
<h3>
Answer: 2.26</h3>
The slope:
m = ( y2 - y1 ) / ( x2 - x1 ) = ( 8 2 ) / ( 3 - 2 ) = 6 / 1
m = 6 ( we have the same slope for AB and A`B` )
AB = √[( 3 - 2 )² + ( 8 - 2 )²] = √37
A`B` = 3.5 √37 = 21.29