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Shkiper50 [21]
3 years ago
13

M∠MNO =____because they are_____ (Fill in the blanks)

Mathematics
2 answers:
xz_007 [3.2K]3 years ago
7 0

Answer:

But then there is no < MNO on the diagram

Elodia [21]3 years ago
5 0

Answer:

No answer is possible from the information provided

Step-by-step explanation:

The is no∠MNO in the image.

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What is 6 13/50 as a decimal
faust18 [17]
6 13/50 as a decimal is 6.26
Hope this helps
4 0
3 years ago
Indentify the terms coefficients, and constants in each expression
Lyrx [107]
4y + 5+ 3y

The coefficients would be 4 and 3 because they have a variable after them and the constant would be 5.
6 0
2 years ago
Read 2 more answers
Hey, Y'all. Was wonderin' bout this pesky question that got me a bit stumped. Thank ya for your help.
puteri [66]
Answer: Choice B) (24,10)

---------------------------------------------------------

Work Shown:

2x - 4y = 8
2( x ) - 4y = 8
2( 3y-6 ) - 4y = 8 ... notice x has been replaced with 3y-6
2(3y)+2(-6) - 4y = 8
6y-12 - 4y = 8
2y-12 = 8
2y-12+12 = 8+12 ... add 12 to both sides
2y = 20
2y/2 = 20/2 ... divide both sides by 2
y = 10

If y = 10, then
x = 3y-6
x = 3*10-6 ... replace y with 10
x = 30-6
x = 24

Put together, the solution is (x,y) = (24,10)
which is why the answer is choice B

As a check, we can plug (x,y) = (24,10) into each equation
x = 3y-6
24 = 3*10 - 6
24 = 30 - 6
24 = 24 ... true equation
and similarly for the second equation as well
2x-4y = 8
2*24 - 4*10 = 8
48 - 40 = 8
8 = 8 ... true equation
Both equations are true when (x,y) = (24,10) so the solution is confirmed
4 0
3 years ago
What are some ways you can make a rectangle 10,000 cm squared (I NEED A AWNSER QUICK)​
lidiya [134]

Answer:

2 longer sides would be 4000 each, so that's 8000 total

and the 2 shorter sides could be 1000 each, so that's 2000 total

8000+2000=10,000

Step-by-step explanation:

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
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