Answer:
x = number of packets of strawberry wafers = 4 packets
y = number of packets of chocolate wafers = 18 packets
Step-by-step explanation:
Devon and his friends bought strawberry wafers for $3 per packet and chocolate wafers for $1 per packet at a carnival. They spent a total of $30 to buy a total of 22 packets of wafers of the two varieties.
Part A: Write a system of equations that can be solved to find the number of packets of strawberry wafers and the number of packets of chocolate wafers that Devon and his friends bought at the carnival. Define the variables used in the equations. (5 points)
Part B: How many packets of chocolate wafers and strawberry wafers did they buy? Explain how you got the answer and why you selected a particular method to get the answer. (5 points)
A.
Let
x = number of packets of strawberry wafers
y = number of packets of chocolate wafers
x + y = 22
3x + y = 30
B. Solve the equation in A using elimination method
x + y = 22 (1)
3x + y = 30 (2)
subtract (1) from (2) to eliminate y
3x - x = 30 - 22
2x = 8
x = 8/2
x = 4
Substitute x = 4 into (1) to solve for y
x + y = 22 (1)
4 + y = 22
y = 22 - 4
y = 18
x = number of packets of strawberry wafers = 4 packets
y = number of packets of chocolate wafers = 18 packets
Answer:
10
Step-by-step explanation:
3^2*2-4(3-1)
3^2=9
9*2
9×2
18
4(3-1)
4(2)
8
18-8
=10
So the final answer is 10
Answer:
it must also have the root : - 6i
Step-by-step explanation:
If a polynomial is expressed with real coefficients (which must be the case if it is a function f(x) in the Real coordinate system), then if it has a complex root "a+bi", it must also have for root the conjugate of that complex root.
This is because in order to render a polynomial with Real coefficients, the binomial factor (x - (a+bi)) originated using the complex root would be able to eliminate the imaginary unit, only when multiplied by the binomial factor generated by its conjugate: (x - (a-bi)). This is shown below:
![(x-(a+bi))*(x-(a-bi))=\\(x-a-bi)*(x-a+bi)=\\([x-a]-bi)*([x-a]+bi)=\\(x-a)^2-(bi)^2=\\(x-a)^2-b^2(-1)=\\(x-a)^2+b^2](https://tex.z-dn.net/?f=%28x-%28a%2Bbi%29%29%2A%28x-%28a-bi%29%29%3D%5C%5C%28x-a-bi%29%2A%28x-a%2Bbi%29%3D%5C%5C%28%5Bx-a%5D-bi%29%2A%28%5Bx-a%5D%2Bbi%29%3D%5C%5C%28x-a%29%5E2-%28bi%29%5E2%3D%5C%5C%28x-a%29%5E2-b%5E2%28-1%29%3D%5C%5C%28x-a%29%5E2%2Bb%5E2)
where the imaginary unit has disappeared, making the expression real.
So in our case, a+bi is -6i (real part a=0, and imaginary part b=-6)
Then, the conjugate of this root would be: +6i, giving us the other complex root that also may be present in the real polynomial we are dealing with.
Answer:
first off, take pic right side up next time. secondly, it's the forth one down
Step-by-step explanation:
Step-by-step explanation:
fill in the missing fraction parts
