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Iteru [2.4K]
2 years ago
10

Write 37 and 2/3% as a fraction in simplest form. URGENT

Mathematics
1 answer:
Tems11 [23]2 years ago
8 0

Answer:

101 2/3

Step-by-step explanation:

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Find the area of the shaded region .
klasskru [66]
Area of the shaded region=area of the greater rectangle- area of the inner rectangle.

Area of a rectangle=length x width.

Area of the greater rectangle=(8x-1)(6x)
area of the inner rectangle=(5x+3)(2x)

Area of the shaded region=(8x-1)(6x)-(5x+3)(2x)
=48x²-6x-(10x²+6x)
=48x²-6x-10x²-6x
=38x²-12x

Answer: the area of the shaded region would be: 38x²-12x
8 0
3 years ago
Write 0.414141414141 as a fraction?
igor_vitrenko [27]

Answer:

The fraction is 41/99.

Step-by-step explanation:

The steps are :

let \: x \: be \: fraction,

x = 0.414141.....

100x = 41.414141.....

100x - x = 41.414141.....  \:  -  \: 0.414141.....

99x = 41

x =  \frac{41}{99}

4 0
3 years ago
What is the midpoint for (-10,3) and (-3,-4)
vova2212 [387]

The midpoint is (-6.5,-0.5)

try using Desmos.com, when dealing with coordinates and functions. It really works!

7 0
2 years ago
How many sides does a regular polygon have if one of it's interior angles measure 162?
julsineya [31]
If you are given the measure of each interior angle (162 degrees) of a regular polygon. how many sides does the polygon have? If you look at a pentagon first where the interior angle is 108 deg then you see. The angle at the center is 180-162=18 deg and 360/18=20 so the polygon has 20 sides.
Hope this helps you !!
8 0
3 years ago
Read 2 more answers
wo point charges lie on the $x$ axis. A charge of +6.24 $\mu C$ is at the origin, and a charge of -9.55 $\mu C$ is at $x$ = 12.0
Andreyy89

Answer:

E_n=34,467,075.42\ N/C

Step-by-step explanation:

<u>Electric Field</u>

The electric field produced by a point charge Q at a distance d is given by

\displaystyle E=K\cdot \frac{Q}{d^2}

Where

K = 9\cdot 10^9\ Nw.m^2/c^2

The net electric field is the vector addition of the individual electric fields produced by each charge. The direction is given by the rule: If the charge is positive, the electric field points outward, if negative, it points inward.

Let's calculate the electric fields of each charge at the given point. The first charge q_1=+6.24\mu C=6.24\cdot 10^{-6}C is at the origin. We'll calculate its electric field at the point x=-3.85 cm. The distance between the charge and the point is d=3.85 cm = 0.0385 m, and the electric field points to the left:

\displaystyle E_1=9\cdot 10^9\cdot \frac{6.24\cdot 10^{-6}}{0.0385^2}

E_1=37,888,345.42\ N/C

Similarly, for q_2=-9.55\mu C=-9.55\cdot 10^{-6}C, the distance to the point is 12 cm + 3.85 cm = 15.85 cm = 0.1585 m. The electric field points to the right:

\displaystyle E_2=9\cdot 10^9\cdot \frac{9.55\cdot 10^{-6}}{0.1585^2}

E_2=3,421,270\ N/C

Since E1 and E2 are opposite, the net field is the subtraction of both

E_n=37,888,345.42\ N/C-3,421,270\ N/C

\boxed{E_n=34,467,075.42\ N/C}

6 0
3 years ago
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