Peter the optician sees 16 patients one Tuesday.4 are neither short nor long sighted, the total number of short sighted patients is 2 more than the total number of long sighted, how many patients can be short sighted, how many long sighted, and how many will require varifocal lenses

Solution:

Number of patients  having neither long nor shortsightedness = 4

Let number of patients having long sightedness be "x"

The total number of short sighted patients is 2 more than the total number of long sighted

Therefore,

Number of patient having shortsightedness = x + 2

Peter the optician sees 16 patients one Tuesday

Total number of patients = 16

Therefore,

x + x + 2 + 4 = 16

2x + 6 = 16

2x = 16 - 6

2x = 10

x = 5

Number of patients having longsightedness = x = 5

Number of patient having shortsightedness = x + 2 = 5 + 2 = 7

Patients requiring varifocal lenses will be = 5 + 7 = 12  

the answer is b: 18

step-by-step explanation:

use the pythagorean theorem to .

a^2+b^2=c^2

plug 18 for "r"

the 2 legs are 18 and 24

18^2+24^2=c^2

324+576=c^2

900=c^2

c=30

30-12=18

18 is the radius

: )

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.