9514 1404 393
Answer:
C) 4
Step-by-step explanation:
The point-slope equation of the line is ...
y -k = m(x -h) . . . . . line with slope m through point (h, k)
y +9 = 3(x -1) . . . . . line with slope 3 through point (1, -9)
For y=0 (the 0 of the function), we have ...
9 = 3(x -1)
3 = x -1
4 = x
The x-intercept is (4, 0).
Answer:
X density = fXpxq and
Y" =InpXq
Now to find Y density FYpyq interms of the density of X we compare the density of X with Y"
fX = In
And PXq =pxq
Thus replacing x with y,
PXq = pyq
(a) Hence the density of Y is FYpyq
(b) at p0, fYpyq =fYp0q= 0
At 5s, FYpyq =5
Answer:
The last one (f) hope u still have time
Answer:
See the attached figure which represents the problem.
As shown, AA₁ and BB₁ are the altitudes in acute △ABC.
△AA₁C is a right triangle at A₁
So, Cos x = adjacent/hypotenuse = A₁C/AC ⇒(1)
△BB₁C is a right triangle at B₁
So, Cos x = adjacent/hypotenuse = B₁C/BC ⇒(2)
From (1) and (2)
∴ A₁C/AC = B₁C/BC
using scissors method
∴ A₁C · BC = B₁C · AC
