Answers7, 11 are linear.
9 is a quadratic. y = 2x^2 - 3
Problem 11First find the slope. Use the first two points to do that.
y2 = 8
y1 = 6
x2 = - 3
x1 = - 1
<u>Sub and solve</u>
y = (y2 - y1) / (x2 - x1)
y = (8 - 6) / (-3 - - 1)
y = 2/- 2
y = - 1
<u>Step 2</u>
Find the y intercept
So far we have
y = - x + b
x = 0
y = 5
5 = 0 *-1 + b
b = 5
<u>Answer</u>
y = -x + 5
Problem 9 is the quadratic.
y = a*x^2 + b
when x = 0
y = - 3
y = ax^2 - 3 Now use any other point to solve for a.
Use (2,5)
5 = a(2)^2 - 3 Add 3 to both sides.
5 + 3 = a*4
8 = 4a Divide by 4
8/4 = a
2 = a
<u>check</u>
Use (-1,1) to check
y = 2x^2 - 3
y = 2(-1)^2 - 3
y = 2*1 - 3
y = 2 - 3
y = - 1 Which is as it should be.
Problem 7 Is linear. Do the same way as 11.
The graph is included below.
See if you can get the equation line.
You should get
y = 2x - 4
3*5=15, you didn’t specify how much chocolate was in the box, use the number of chocolate in the box and subtract it by 15, whatever number you get from that problem should be how much she’ll have remaining.
The constant rate of change is $0.30
One ear of corn costs $0.30
First, convert both fractions into improper fractions:
-2 and 1/4= -9/4
-4 and 1/2= -9/2
Now we can find the product!
-9/4 x -9/2= 81/8
(Remember that when you multiply two negatives, the product turns positive)
And because 81/8 cannot be reduced, to find your final answer, just covert 81/8 into a mixed fraction!
81/8= 10 and 1/8
Hope this helps!