Answer:
all of them
Step-by-step explanation:
:)
yepppp I'm pretty surr
<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
3(x+2) < 10 //// that should be right!!
Answer:
6
Step-by-step explanation:
We can fill in what we know which is the length and perimeter.
82=2(35)+2(w)
Multiply 35 x 2 = 70
82-70= 12
12/2=6
Answer:
Step-by-step explanation:
(0, -9) and (3,3)
(3+9)/(3-0) = 12/3 = 4
y + 9 = 4(x - 0)
y + 9 = 4x - 0
y = 4x - 9