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:
-24+12(d-3)+22=-24+34(d-3)
10(d-3)
10d=-30
d=-30/10
d=3
Answer:
(x-2)²/100 + (y+3)²/64 = 1
Step-by-step explanation:
C (2,-3): h=2 k=-3
semimajor axis (CV): a=12-2=10
center-foci: c=8-2=6
semi minor axis: b² = a²-c²= 100 - 36 = 64
equation: (x-h)²/a² + (y-k)²/b² = 1
(x-2)²/100 + (y+3)²/64 = 1
7
5 times 7
equals 35
then you add 15
and get 50
