Astamp collector has some 15¢ stamps and some 20¢ stamps. the number of 15¢ stamps is eight less
than three times the number of 20¢ stamps. the total value is $4. find the number of each type of
stamp in the collection.

Step-by-step explanation:

Let variable x represent the 15¢ stamps, and variable y  represent 20¢ stamps.

x= 3y-8

0.15x+0.20y = 4

Replace x in the second equation by (3y-8)

0.15(3y-8) + 0.20y = 4

0.45y - 1.2 +0.20y = 4

Collect like terms;

0.45y+0.20y = 4+1.2

0.65y = 5.2

Divide both sides by 0.65 to solve for y;

y= 5.2/0.65

y = 8

Replace in equation " x=3y-8 " to solve for x;

x = (3*8) -8

x=24 - 8

x = 16

Therefore, the 15¢ stamps are 16 in number, and the 20¢ stamps are 8 in number.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

option c) graph of line going through 1, 4 and 3, 10

step-by-step explanation:

let y equal to margo's coin collection

let x equal to number of years

at the begining x=0, y=1 → point=(x,y)→point=(0,1)

she adds 3 coins per year, then y=1+3x

at x=1, y=1+3(1)=1+3→y=4→point=(x,y)→point=(1,4)

at x=3, y=1+3(3)=1+9→y=10→point=(x,y)→point=(3,10)

the graph of line going through (1, 4) and (3, 10), then the option c is the correct.