To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.
jk=ac and j+k=b
The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...
2k^2-5k-18=0
2k^2+4k-9k-18=0
2k(k+2)-9(k+2)=0
(2k-9)(k+2)=0
so k=-2 and 9/2
k=(-2, 4.5)
Answer:
First option: cos(θ + φ) = -117/125
Step-by-step explanation:
Recall that cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)
If sin(θ) = -3/5 in Quadrant III, then cos(θ) = -4/5.
Since tan(φ) = sin(φ)/cos(φ), then sin(φ) = -7/25 and cos(φ) = 24/25 in Quadrant II.
Therefore:
cos(θ + φ) = cos(θ)cos(φ) - sin(θ)sin(φ)
cos(θ + φ) = (-4/5)(24/25) - (-3/5)(-7/25)
cos(θ + φ) = (-96/125) - (21/125)
cos(θ + φ) = -96/125 - 21/125
cos(θ + φ) = -117/125
Answer:
Chukliya bhaiya iss chiz khi toh mere ko zarurat thi zindagi mein!!(╥﹏╥)(╥﹏╥)
14 out of 24 can be simplified to 7/12 which is 58.3%
