Well
It is in the form of a/b + b/a = (a sqr + b sqr)ab
Substitute
(Sin sqr theta + cos sqr theta) +2 cos theta + 1. 2 + 2cos theta
———————————————————————— = ————————-
Sin theta(1+cos theta). Sin theta (1 + cos theta)
2(1+ cos theta)/sin theta(1 + cos theta). {1 + cos theta cancelled}
2/sin theta. Hope it helps :}
Hello.
Since 3 < pi < 4,
√9 < pi √16
In fact, since pi^2 = 9.86,
<span>√9 < pi < √10.
Which means the you</span><span> can find pi between square roots √9 and √10.
</span>
Have a nice day
Answer:
Step-by-step explanation:
For this equation: a=3, b=4, c=2
3x2+4x+2=0
Step 1: Use quadratic formula with a=3, b=4, c=2.
x= −b±√b2−4ac / 2a
x= −(4)±√(4)2−4(3)(2) / 2(3)
x= −4±√−8 / 6
Answer:
No real solutions.
Answer:
x-coordinate = 0
Step-by-step explanation:
(-10,1) , (5,-5)
m = (y₂ - y₁)/(x₂ - x₁) = ( -5 - 1 ) / ( 5 - [-10] )
= -6 / ( 5 + 10)
= -6/15 = -2/5
y - y₁ = m (x - x₁)
y - 1 = (-2/5) ( x - [-10] )
y - 1 = (-2/5) ( x + 10 )
5 * (y -1) = -2 (x + 10)
5y - 5 = -2x -20
2x + 20 + 5y -5 = 0
2x + 5y + 15 = 0
Another point (a, -3) in on this line
2 * a + 5 * (-3) = -15
2a -15 = -15
2a = -15 +15
2a = 0
a = 0
x-coordinate = 0
