<h3>Jason bought 20 stamps of $0.41 each and 8 postcards of $0.26 each.</h3>
<em><u>Solution:</u></em>
Let stamps be s and postcards be p
Given that,
The number of stamps was 4 more than twice the number of postcards
s = 4 + 2p -------- eqn 1
Jason bought both 41-cent stamps and 26-cent postcards and spent $10.28
41 cent = $ 0.41
26 cent = $ 0.26
Therefore,

0.41s + 0.26p = 10.28 --------- eqn 2
Substitute eqn 1 in eqn 2
0.41(4 + 2p) + 0.26p = 10.28
1.64 + 0.82p + 0.26p = 10.28
1.08p = 10.28 - 1.64
1.08p = 8.64
Divide both sides by 1.08
p = 8
Substitute p = 8 in eqn 1
s = 4 + 2(8)
s = 4 + 16
s = 20
Thus Jason bought 20 stamps and 8 post cards
Answer:
x=3
Step-by-step explanation:
x^2 = 6x - 9
Subtract 6x from each side
x^2 -6x = -9
Add 9 to each side
x^2 -6x+9 = 0
Factor
what 2 numbers multiply to 9 and add to -6
-3*-3 = 9
-3+-3 =-6
(x-3)(x-3) = 0
Using the zero product property
x-3 =0 x-3=0
x=3 with multiplicity 2
first combine like terms so 2n + 7n which is 9n
so 9n + 7 = 13 + 3n + 8n
then combine on other side so 9n + 7 = 13 + 11n
then subtract 7 to 13 so 6
sp 9n + 6 = 11n
then subtract 9 to the other side so 2n
6 = 2n then divide all by 2
n= 3
Step-by-step explanation:
Answer:
2 units
Step-by-step explanation:
none