Peter the optician sees 16 patients one Tuesday. 4 are neither short nor long sighted. The total number of short sighted patients is two more than the total number of long sighted patients. How many patients can be short sighted, how many can be long sighted and how many could require varifocal lenses?

No of patients having longsightedness =  5

No of patient having shortsightedness = 7

Patients requiring varifocal lenses will be  =12                  

Step-by-step explanation:

Total no of patients = 16

As

No of patients  having neither long nor shortsightedness = 4

Let

No of patients having longsightedness = x

So

No of patient having shortsightedness =

Equation becomes

Now

No of patients having longsightedness = x= 5

No of patient having shortsightedness =      =

                                                                 = 7

Patients requiring varifocal lenses will be = 5+7

                                                                         = 12

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²