Y= x^2-4x+3 in vertex form
1 answer:
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Answer:
Step-by-step explanation:
Let the consecutive even numbers be x,x+2,x+4, x+ 6, x+ 8 and x+10
So, the given condition is:
x+x+2+x+4+x+6+x+8+x+10 = 270
6x + 30 = 270
Subtracting both sides by 30
6x = 270 - 30
6x = 240
Dividing both sides by 6
x = 40
So,
First Number = x = 40
Second Number = x+2 = 40+2 = 42
Third Number = x+4 = 40+4 = 44
Fourth Number = x+6 = 40+6 = 46
Fifth Number = x+8 = 40+8 = 48
Sixth Number = x+10 = 40+10 = 50
Answer:
Step-by-step explanation:
Begin by squaring both sides to get rid of the radical. Doing that gives you:
Now use the Pythagorean identity that says
and make the replacement:
. Now move everything over to one side of the equals sign and set it equal to 0 so you can factor:
and then simplify to
Factor out the common cos(x) to get
and there you have your 2 trig equations:
cos(x) = 0 and 1 - cos(x) = 0
The first one is easy enough to solve. Look on the unit circle and see where, one time around, where the cos of an angle is equal to 0. That occurs at
The second equation simplifies to
cos(x) = 1
Again, look to the unit circle and find where the cos of an angle is equal to 1. That occurs at π only.
So, in the end, your 3 solutions are