Find four numbers that form a geometric progression such that the third term is greater than the first by 12 and the fourth is greater than the second by 36.

5 , 4.5, 13.5 and 40.5

Step-by-step explanation:

Since the numbers are in geometric progression, their form is essentially:

a, ar, ar^2 and ar^3

Where a and r are first term and common ratio respectively.

From the information given in the catalog:

Third term is greater than the first by 12 while fourth is greater than second by 36.

Let’s now translate this to mathematics.

ar^2 - a = 12

ar^3 - ar = 36

From 1, a(r^2 - 1) = 12 and 2:

ar(r^2 - 1) = 36

From 2 again r[a(r^2 -1] = 36

The expression inside square bracket looks exactly like equation 1 and equals 12.

Hence, 12r = 36 and r = 3

Substituting this in equation 1,

a( 9 - 1) = 12

8a = 12

a = 12/8 = 1.5

Thus, the numbers are 1.5, (1.5 * 3) , (1.5 * 9), (1.5 * 27) = 1.5 , 4.5, 13.5 and 40.5

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10