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Korolek [52]
3 years ago
10

Need Help ASAP!!

Mathematics
1 answer:
egoroff_w [7]3 years ago
4 0
Bxh

14x15= 210
Hope this help
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What is the area of a square with a side length of 10cm?
natita [175]

Answer:

100cm

Step-by-step explanation:

10*10

4 0
3 years ago
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PLEASEEEEEEEEEE HELPPPPPPPPPPP
ankoles [38]

Answer:

I would just guess, but if i were to actually think about it, i would think C. I hope it helps, and blame it on me if you fail.

Step-by-step explanation:

8 0
3 years ago
Find the value of x for mAB=34 and mCD=25 (The figure is not drawn to scale.)
Alexus [3.1K]
<h3>Answer:   29.5 degrees</h3>

==============================================

Explanation:

Add up the arc measures and divide by 2 to get the angle formed by the two chords.

angle AOB = (arc AB + arc CD)/2

angle AOB = (34 + 25)/2

angle AOB = 59/2

angle AOB = 29.5 degrees

3 0
3 years ago
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A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and
nasty-shy [4]

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = \rho

case (1)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

so here

l = \frac{8 \lambda _1}{2}      ..............1

l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}

and we know velocity is express as

velocity = frequency × wavelength   .....................2

\sqrt{\frac{Tension}{mass\ per\ unit \length }}   =   f × \lambda_1

here tension = mg

so

\sqrt{\frac{mg}{\mu}}   =   f × \lambda_1     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

l = \frac{10 \lambda _1}{2}      ..............4

l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}

when block is immersed

equilibrium  eq will be

Tenion + force of buoyancy = mg

T + v × \rho × g = mg

and

T = v × \rho - v × \rho × g    

from equation 2

f × \lambda_2 = f  × \frac{1}{5}  

\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}     .......................5

now we divide eq 5 by the eq 3

\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}

solve irt we get

1 - \frac{\rho _{stone}}{\rho _{water}}  = \frac{16}{25}

so

relative density \frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

3 0
4 years ago
Oints (2,5 and 8,4
Zanzabum
The slope for the question is -1.5
7 0
2 years ago
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