Let's start by writing a system of linear equations:
c -> cookies
cb -> candy bars
(You can use any abbreviations to your preference)
Abby:
4 cookies
3 candy bars
$10.25 per bag
The equation would be:
4c+ 3cb = $10.25
Marissa:
2 cookies
7 candy bars
$14.75 per bag
The equation would be:
2c + 7cb = $14.75
So our linear equation system would be:
<span>4c+ 3cb = $10.25
</span><span>2c + 7cb = $14.75
I would try to get rid of one variable so I can solve for the other variable. In this case, it is easier to get rid of c since I can multiply the second equations by 2. Then it would subtract the two equations.
(2c + 7cb = $14.75) 2 = 4c + 14 cb = $29.50
4c + 3cb = $10.25
- 4c+14 cb = $29.50 (4c would get canceled.)
---------------------------------
-11 cb = - $19.25 (Divide by -11 to solve for cb)
</span> --------- -------------
-11 -11
cb = $1.75
Now we know cb (candy bar) cost, we would substitute this value into cb into one of the equations. It doesn't matter which equation you put it in. I will substitute it in the first equations.
4c + 3 (1.75) = $10.25
4c + 5.25 = $10.25 (Multiply 3 by 1.75)
-5.25 -5.25 (Subtract 5.25 on both sides)
4c = 5 (Divide by 4 on both sides to get c)
---- ---
4 4
c= 1.25
Check the work:
4(1.25) + 3(1.75)
= $10.25
2(1.25) + 7(1.75)
= $14.75
Total cost:
cookies = $1.25
candy bars = $ 1.75
Hope this helps! :)
Answer:
- Commutative Property of Addition.-
- 2 + 3 = 3 + 2
- Commutative Property of Multiplication.- ex. 5 × 7 = 7 × 5
- Associative Property of Addition.- ex.(2 + 3) + 6 = 2 + (3 + 6)
- Associative Property of Multiplication.- ex. (7 × 3) × 10 = 7 × (3 × 10)
- Distributive Properties of Addition Over Multiplication- ex. 2 × (2 + 8) = 2 × 2 + 2 × 8
- (2 + 8) × 10 = 2 × 10 + 8 × 10
- The reciprocal of a non zero real number a is 1/a.- ex. reciprocal of 5 is 1/5 and 5 × (1/5) = 1
Step-by-step explanation:
Answer:
nine hours
Step-by-step explanation:
48 divided by 6 is 8 and 8 times 9 is 72
