Answer:
There are no real solutions. The complex solutions are ...
(x, y) = (2.5-i√1.75, -2.5+i√1.75) or (2.5+i1.75, -2.5-i1.75)
Step-by-step explanation:
Subtract the first equation from the second.
y -y = (x^2 -6x +8) -(-x)
0 = x^2 -5x +8
Rewrite to facilitate completing the square:
x^2 -5x = -8
x^2 -5x +6.25 = -1.75 . . . . . add 6.25 to complete the square
(x -2.5)^2 = -1.75 . . . . . write as a square
x -2.5 = ±i√1.75 . . . . . . take the square root
x = 2.5 ± i√1.75 . . . . . . add 2.5; y is the opposite of this
Solutions are ...
- x = 2.5 +i√1.75, y = -2.5 -i√1.75
- x = 2.5 -i√1.75, y = -2.5 +i√1.75
_____
The graph shows the equations have no point of intersection, meaning there are no real solutions. The complex solutions are shown above.
Answer:
B) 6(x-5) (x+5)
Step-by-step explanation:
Common factor.
6
(
^2
−
25)
Use the sum-product pattern.
6
(
^2+5−5−25).
Common factor from the two pairs.
6((+5)−5(+5)).
Rewrite in factored form.
6
(
−
5
)
(
+
5).
Answer:
7 is the y-intercept
slope is 2/1
Step-by-step explanation:
Answer:
See the step by step solution
Step-by-step explanation:
Consider a more general case of the form
. We want to express it as a sine term only by using the following formula 
The trick consists on find a number
such that
. But, note that since
it is most likely that
. Then, we will use the following equations:
![\cos(\beta) =\frac{a}{\sqrt[]{a^2+b^2}=A,\sin(\beta) =\frac{b}{\sqrt[]{a^2+b^2}=B](https://tex.z-dn.net/?f=%5Ccos%28%5Cbeta%29%20%3D%5Cfrac%7Ba%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DA%2C%5Csin%28%5Cbeta%29%20%3D%5Cfrac%7Bb%7D%7B%5Csqrt%5B%5D%7Ba%5E2%2Bb%5E2%7D%3DB)
Then, problem turns out to be
,
where ![\beta = \tan^{-1}(\frac{B}{A})= \tan^{-1}(\frac{b}{a}). So, note that the frequency is the same. Then, a) Taking a=4 and b= 15 we have that 4 sin 2πt + 15 cos 2πt= [tex]\sqrt[]{15^2+4^2}\sin(2\pi t + \tan^{-1}\frac{15}{4})](https://tex.z-dn.net/?f=%5Cbeta%20%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7BB%7D%7BA%7D%29%3D%20%5Ctan%5E%7B-1%7D%28%5Cfrac%7Bb%7D%7Ba%7D%29.%20So%2C%20note%20that%20the%20frequency%20is%20the%20same.%20%3C%2Fp%3E%3Cp%3EThen%2C%20%3C%2Fp%3E%3Cp%3Ea%29%20Taking%20a%3D4%20and%20b%3D%2015%20we%20have%20that%204%20sin%202%CF%80t%20%2B%2015%20cos%202%CF%80t%3D%20%5Btex%5D%5Csqrt%5B%5D%7B15%5E2%2B4%5E2%7D%5Csin%282%5Cpi%20t%20%2B%20%5Ctan%5E%7B-1%7D%5Cfrac%7B15%7D%7B4%7D%29)
b) The amplitude is
. Since
the frequency is 1 Hz. The period is given by \frac{2\pi}{2\pi} = 1.
c) The function's graph is attached