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.
10^1 = 10
10^2 = 100 {10 x 10 = 100}
10^3 = 1000 {10 x 10 x 10 = 1,000}
10^4 = 10000 {10 x 10 x 10 x 10 = 10,000}
The exponent represents how many zeros will be used in the answer. For example, 10^6 would include 6 zeros in it’s answer; 1,000,000.
2/4? 4/8? .5? All mean 1/2. Also one half.
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The value should the equipment be recorded in aaron company's records is <span>90,000</span>
