Answer:
Step-by-step explanation:
product of slopes of perpendicular lines=-1
(t-5)/(3+4)×(2-3)/(-4-1)=-1
(t-5)/7×(-1/-5)=-1
(t-5)/35=-1
t-5=-1×35=-35
t=-35+5
t=-30
2.
slopes of parallel lines are equal.
(-2+3)/(t-4)=(-1-4)/(4+2)
1/(t-4)=-5/6
t-4=-6/5
t=4-6/5=(20-6)/5=14/5
3.
x>0,y<0
so P lies in4th quadrant.
except cos and sec all are negative.
so only cos and sec are positive.
Hi, thank you for posting your question here at Brainly.
This problem could be solved using the Law of Cosines, since we are given with the three sides. The equation goes,
b^2 = a^2 + c^2 - 2accosB. Substituting,
24^2 = 7^2 + 25^2 - 2(7)(25)cosB
Then B = 73.7 degrees. Therefore sinB or sin(73.7) = 0.96.
Answer:
3 x^2 - x + -1
Step-by-step explanation:
Simplify the following:
-(4 x - 2 x^2 - 3) + x^2 + 3 x - 4
Factor -1 out of -2 x^2 + 4 x - 3:
--(2 x^2 - 4 x + 3) + x^2 + 3 x - 4
(-1)^2 = 1:
2 x^2 - 4 x + 3 + x^2 + 3 x - 4
Grouping like terms, 2 x^2 + x^2 + 3 x - 4 x - 4 + 3 = (x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3):
(x^2 + 2 x^2) + (3 x - 4 x) + (-4 + 3)
x^2 + 2 x^2 = 3 x^2:
3 x^2 + (3 x - 4 x) + (-4 + 3)
3 x - 4 x = -x:
3 x^2 + -x + (-4 + 3)
3 - 4 = -1:
Answer: 3 x^2 - x + -1
