To see if the point is on the line, plug in one of the point values to see if it equals out.
y = 4x - 2 I will plug in 1 for x to see if y does = 3
y = 4(1) - 2
y = 4 - 2
y = 2 so (1, 2)
Dinesh is not correct because the point on the line when x = 1 is (1, 2) not (1, 3)
By adding, subtracting, multiplying or dividing numbers
ANSWER:
Let t = logtan[x/2]
⇒dt = 1/ tan[x/2] * sec² x/2 × ½ dx
⇒dt = 1/2 cos² x/2 × cot x/2dx
⇒dt = 1/2 * 1/ cos² x/2 × cosx/2 / sin x/2 dx
⇒dt = 1/2 cosx/2 / sin x/2 dx
⇒dt = 1/sinxdx
⇒dt = cosecxdx
Putting it in the integration we get,
∫cosecx / log tan(x/2)dx
= ∫dt/t
= log∣t∣+c
= log∣logtan x/2∣+c where t = logtan x/2
We know that
points are
x intercept
A (4,0)
y intercept
B (0,11)
step 1
find the equation of a line
m=(y2-y1)/(x2-x1)--------> m=(11-0)/(0-4)------> m=-11/4
with m and the point B (0,11)
y-y1=m*(x-x1)
y-11=(-11/4)*(x-0)---------> y=-(11/4)x+11
the answer is
<span>the formula of the function is
</span>y=-(11/4)x+11
see the attached figure
Answer:
y - 1 = 0
Step-by-step explanation:
move constant to the left by adding its opposite to both sides y - 1 = 1 - 1
the sum two opposites equals 0
y =1