Y = x^2 - 8x + 5 Put brackets around the first two terms.
y = (x^2 - 8x ) + 5 Add (8/2)^2 on the inside of the brackets. Translated into mathese that means take 1/2 of the linear term (8x), drop the x and square the rest
.
(1/2 8)^2 = 4^2 = 16
y = (x^2 - 8x + 16) - 5 What you put inside the brackets, you must subtract outside the brackets.
y = (x^2 - 8x + 16) - 5 - 16 Combine like terms.
y = (x^2 - 8x + 16) - 21 Show what is inside the brackets as a square.
y = (x - 4)^2 - 21 <<<<<<<<< Answer.
Answer:
Step-by-step explanation:
Later on in the course, I hope you are told that answers should not rely on diagrams.
Since you have to answer the question somehow, the answer (directly) is BCA. Since this is an appearance question (that's what the answer looks like), you can only state the answer, There really (in this case) is no what to offer a proof).
I would use a proportion
45students/60% = Xstudents/100% (cross multiply)
x=75 students
Answer: (c)
Step-by-step explanation:
Given
The function is ![y=-x^2+6x-4](https://tex.z-dn.net/?f=y%3D-x%5E2%2B6x-4)
Range of a function describes the values y can take
from the graph it is clear that the maximum value of y is 5
![\therefore y\leq5](https://tex.z-dn.net/?f=%5Ctherefore%20y%5Cleq5)
Hence. option (c) is correct.
The answer is B because the first number in those points don’t repeat but all the other ones have the first number repeating