Answer:
a-bi
Step-by-step explanation:
If a quadratic equation lx^2+mx+n=0 has one imaginary root as a+bi then the other root is the conjugate of a+bi = a-bi
Because we have l, m and n are real numbers and they are the coefficients.
Sum of roots = a+bi + second root = -m/l
When -m/l is real because the ratio of two real numbers, left side also has to be real.
Since bi is one imaginary term already there other root should have -bi in it so that the sum becomes real.
i.e. other root will be of the form c-bi for some real c.
Now product of roots = (a+bi)(c-bi) = n/l
Since right side is real, left side also must be real.
i.e.imaginary part =0
bi(a-c) =0
Or a =c
i.e. other root c-bi = a-bi
Hence proved.
I believe it’s 3, because the radius is half of the diameter
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Answer:
.
Step-by-step explanation:
(5x^2 + 2y^2) * (3x^2 - 7y^2)
= 15x^4 + 6x^2y^2 - 35x^2y^2 - 14y^4
= 15x^4 - 29x^2y^2 - 14y^4
=
Hope this helps!
Answer:
The cosine function is f(t) = -30sin(π÷3)t + 20
Step-by-step explanation:
Given : The temperature of a chemical reaction ranges between −10 degrees Celsius and 50 degrees Celsius. The temperature is at its lowest point when t = 0, and the reaction completes 1 cycle during a 6-hour period.
To find : What is a cosine function that models this reaction?
General form of cosine function is f(x) = A cos(Bx)+C
Where A is the amplitude
B=2π÷Period
C is the midline
Now, We have given
The temperature of a chemical reaction ranges between −10 degrees Celsius and 50 degrees Celsius.
A is the average of temperature,
i.e,
A=(-10-50)÷2 = -30
Period of 1 cycle is 6 hour
So,
B = 2π÷6 = π÷3
The temperature is at its lowest point when t = 0 and we know lowest point is -10
So,
f(t) = A cos t + C
-10 = -30 cos 0 + C
Therefore, C = 20
Substituting the values, we get
The cosine function is f(t) = -30sin(π÷3)t + 20
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