I'm guessing your problem is this:
y³ - 9y² + y - 9 = 0
right?
In solving this problem, I recommend doing this:
y³ - 9y² + y - 9 = 0
Factor out a y² from the first two numbers in the problem:
y²(y - 9) + (y - 9) = 0
Separate the parentheses which means y - 9 goes on one side. The y² added a one since it came from the + 1 in the middle of expression. When you're separating parentheses like this you just take the outside numbers and combine them together. Since + 1 came from the outside of the (y - 9) and y² also was sitting on the outside of (y - 9) combine them to make y² + 1. Like this:
(y² + 1)(y - 9) = 0
Now separate your two parentheses to two separate problems:
(y² + 1) = 0 and (y - 9) = 0
Now you're y² + 1 will equal:
y² = -1
y = √-1 <-- This number doesn't exist so it will be an imaginary number (i). If you guys didn't learn that in your class I recommend just leaving it as i for that part.
Now solve y - 9 = 0:
y = 9 <-- Since we added nine to both sides to get this.
So you're final answer should be y = i and 9
Answer: To determine the charge on an object, determine the number of excess protons or excess electrons. Multiply the excess by the charge of an electron or the charge of a proton - 1.6 x 10-19 C.
Step-by-step explanation:
A translation
(I'm assuming that this isn't multiple choice since you didn't include the answer choices.)
The unit rate for the graph above is 60 heartbeats per minute. Half of 2 is 1 so go to the 1 min. mark and go up until you reach the line graph and it says 60. Do the same for the 2 min. mark and go up until you reach the line graph and it'll say 120. 60 ÷60=1 and 120 ÷ 60= 2
Answer:
Step-by-step explanation:
Required to prove that:
Sin θ(Sec θ + Cosec θ)= tan θ+1
Steps:
Recall sec θ= 1/cos θ and cosec θ=1/sin θ
Substitution into the Left Hand Side gives:
Sin θ(Sec θ + Cosec θ)
= Sin θ(1/cos θ + 1/sinθ )
Expanding the Brackets
=sinθ/cos θ + sinθ/sinθ
=tanθ+1 which is the Right Hand Side as required.
Note that from trigonometry sinθ/cosθ = tan θ
