No the corresponding angles are not congruent, because the angle measures on the smaller figure are 90, 90, 137, and 43, while the larger figure has angle measures of 90, 90, 136, 44. that is why the following figures are not congruent.

3 0

3 years ago

Solve the system algebraically. Check your work.

Luden [163]

answer.

Answer:

x=2 and y=0 is the required result.

Step-by-step explanation:

We have been given system of equations:

5x+2y=105x+2y=10 (1)

And 3x+2y=63x+2y=6 (2)

We will use elimination method:

Multiply 1st equation by 3 and 2nd equation by 5 we get:

15x+6y=3015x+6y=30 (3)

15x+10y=3015x+10y=30 (4)

Now subtract (4) from (3) we get:

-4y=0−4y=0

y=0y=0

Now, put y=0 in (1) equation:

5x+2(0)=105x+2(0)=10

5x=105x=10

x=2x=2

Hence, x=2 and y=0

8 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 $\frac{MR^{2} }{8}$ .

Now using perpendicular axis theorem, Ix = Iy = $\frac{MR^{2} }{16}$ = $\frac{ρπR^{4} }{16}$ .

For Radius of gyration K, equate MK2 = MR2/16, K= R/4.

3 0

3 years ago