Altho' I can easily guess what you're supposed to do here, I must point out that you haven't included the instructions for this problem.
I'll help you by example. Let's look at the first problem:
"Evaluate 6(z-1) at z-4."
Due to "order of operations" rules, we must do the work inside the parentheses FIRST. Replace the z inside (z-1) with "-4". We obtain
6(-4-1) = 6(-5) = -30 (answer.)
Your turn. Try the next one. If it's unclear, as questions.
Answer:
Y=-1/3x-1
Step-by-step explanation:
Y-0=-1/3(x+3)
M = -2 and b = 3
so equation
y = -2x + 3
hope it helps
Let P = number of coins of pennies (1 penny = 1 cent)
Let N = number of coins of nickels (1 nickel = 5 cents)
Let D = number of coins of dimes (1 dime = 10 cents)
Let Q = number of coins of quarters (1 quarter = 25 cents)
a) P + N + D + Q = 284 coins, but P = 173 coins, then:
173 + N + D + Q =284 coins
(1) N + D + Q = 111 coins
b) D = N + 5 OR D - N =5 coins
(2) D - N = 5 coins
c) Let's find the VALUE in CENTS of (1) that is N + D + Q = 111 coins
5N + 10D + 25 Q = 2,278 - 173 (1 PENNY)
(3) 5N + 10D + 25Q = 2105 cents
Now we have 3 equation with 3 variables:
(1) N + D + Q = 111 coins
(2) D - N = 5 coins
(3) 5N + 10D + 25Q = 2105 cents
Solving it gives:
17 coins N ( x 5 = 85 cents)
22 coins D ( x 10 = 220 cents)
72 coins D ( x 25 = 1,800 cents)
and 173 P,
proof:
that makes a total of 85+2201800+172 =2,278 c or $22.78
