Answer:
5
is the answer dont over think it and try plz bc its very easy
(2y² + 7y + 11) - (8y² - 5y + 7) you can distribute the negative sign to the second expression and then combine the like terms
new expression: (2y² + 7y + 11) (-8y² + 5y - 7)
2y² and -8y² equals -6y²
7y and 5y equals 12y
11 and -7 equals 4
The answer is -6y² + 12y + 4
Answer:
Divide by 2
q^2+4q=3/2
q^2+4q(4/2)^2=3/2+(4/2)^2
(q+4/2)^2=3/2+16/4
taking the square root of both side
√(q+4/2)^2=√(3/2+16/4)
Note that the square will cancel the square root then you will take LCM on the right hand side
q+4/2=√6+16/4
q+4/2=√22/4
q= -4/2+-√22/4
q=(-4+_√22/4)
Answer:
7
Step-by-step explanation:
There is a point of inflection at each point where the graph crosses the line y=2. There are 7 points of inflection shown.
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A point of inflection is where the graph changes from being concave downward to concave upward, or vice versa. For a sine function, that is everywhere the function crosses its midline.
