We can use elimination for these set of systems.
First, we need to set up our variables.
Belts=b
Hats=h
Now, the situation is 6 belts and 8 hats for $140. The situation after is 9 belts and 6 hats for $132.
Let’s set up our system of equations.
6b+8h=140
9b+6h=132
We need to eliminate a variable. Since b has coefficients of 6 and 9, we can easily eliminate b by multiplying the top equation by 3 and the bottom by -2.
18b+24h=420
-18b-12h=-264
Now let’s add.
12h=156
Let’s divide to get h by itself.
156/12=13=h
So a hat costs $13. We need to put in 13 for one of the equations so we can find the cost of a belt.
9b+6(13)=132
9b+78=132
We need b by itself.
9b=54
54/9=6
Belts are $6
We can also use the first equation to check our answers.
6(6)+8(13)
36+104
140.
So, the price of a belt is $6 while the price of a hat is $13.
Answer:
48
Step-by-step explanation:
pq
substitute values
(8)(6)
8*6 = 48
Answer:
The domain is 4, range is 5
Step-by-step explanation:
Answer:
<h2>x = 10.3</h2>
Step-by-step explanation:
Use sine.
We have:
<em>look at the table in the picture</em>
Substitute:
<em>multiply both sides by 16</em>
7.66. If 100%=1, then 700%= 7. Then if 10%=.1 then 60%=.6. And lastly if 1%=.01 then 6%=.06. Add them all up: 7+.6+.06= 7.66. HTH
