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Schach [20]
3 years ago
9

Scientist Asterix uses 19 base number system. Which

Mathematics
2 answers:
belka [17]3 years ago
4 0

Answer:

A. I

Step-by-step explanation:

first 10 symbols are digits from 0 to 9

then next 9 symbols are alphabets from A to I

so the largest symbol is I

alex41 [277]3 years ago
3 0

19 base number system has 19 symbols:

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, G, H, I

The largest symbol is I

Answer:

<h3>                 A. I</h3>
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Find the area of the shaded region.<br> 8 in<br> 24 in
zloy xaker [14]

Answer:

96 in^2

Step-by-step explanation:

area of rectangle: 24*8=192

Area of triangle: (1/2)*base*height

= .5 * 8 * 24

=192/2

=96

area of shaded=rectangle - triangle

= 192-96

which is also just 192/2

so it is 96

4 0
3 years ago
Two particles are fixed to an x axis: particle 1 of charge -1.00 x 10-7 C is at the origin and particle 2 of charge +1.00 x 10-7
r-ruslan [8.4K]
The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint. 

The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
</span>
Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
6 0
3 years ago
From a cuboidal cardboard box of dimensions 80cm×40cm×15cm,6 circles of radius 7 cm were cut from the front face.Find the surfac
Irina-Kira [14]

Answer:

9076 cm²

Hope my answer helped you

6 0
2 years ago
find the volume of a right circular cone that has a height of 4.2 m and a base with a radius of 3.4 m. Round your answer to the
lianna [129]

Answer: 50.8

Step-by-step explanation:

8 0
3 years ago
Why does (x-2)^2+9=0 have no real solutions?
alex41 [277]

Answer:

see explanation

Step-by-step explanation:

Given

(x - 2)² + 9 = 0 ( subtract 9 from both sides )

(x - 2)² = - 9  ( take the square root of both sides )

x - 2 = ± \sqrt{-9}

\sqrt{-9} yields no real solutions

There is no real number when multiplied by itself gives - 9

In fact the solutions to \sqrt{-9} are said to be imaginary, that is ± 3i

where i = \sqrt{-1}

If interested investigate further.

7 0
3 years ago
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