Please help with my algebra 1

Mathematics

1 answer:

Nitella [24]3 years ago

6 0

Answer:

the answer is c hope this helps

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What are the vertex, focus, and directrix of the parabola with the equation x2+8x+4y+4=0

Novay_Z [31]

x2 +8x +4y +4 = 0
4y= -x2 -8x -4y = -.25*x^2 -2x -1
a = -.25b = -2c = -1
x position of vertex:
h = -b / 2a
h = 2 / 2*-.25h = 2 / -.5h = -4
y position of vertex:
k = ah^2 + bh + ck = -.25*-4^2 + -2*-4 + -1k = -4 +8 -1k = 3
VERTEX = (-4, 3)**************************************************************************
x value of focus =x value of vertex = -4
y value of focus =(1 (-b^2 -4ac)) / 4a

a = -.25 b = -2 c =-1
y value = (1 (-4 -4*-.25*-1)) / 4*-.25
y value = (1 (-4 -4*-.25*-1)) / -1
y value = (1 -4 +1) / -1y value = (-2 / -1)y value = 2
focus value = (-4, 2)
Answer is the last one.

7 0

3 years ago

A lamina with constant density rho(x, y) = rho occupies the given region. Find the moments of inertia Ix and Iy and the radii of

jenyasd209 [6]

Answer:

Ix = Iy = $\frac{ρπR^{4} }{16}$

Radius of gyration x = y = $\frac{R}{4}$

Step-by-step explanation:

Given: A lamina with constant density ρ(x, y) = ρ occupies the given region x2 + y2 ≤ a2 in the first quadrant.

Mass of disk = ρπR2

Moment of inertia about its perpendicular axis is $\frac{MR^{2} }{2}$ . Moment of inertia of quarter disk about its perpendicular is $\frac{MR^{2} }{8}$ .