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4 years ago

One can use two-dimensional objects to build three-dimensional objects? True or False?

Mathematics

2 answers:

umka21 [38]4 years ago

8 0

One can use two-dimensional objects to build three-dimensional objects?

True

Harlamova29_29 [7]4 years ago

3 0

Answer:

TRUE

Step-by-step explanation:

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If the value of a number in decimal system is 955, find its value in quinary system and in binary system.

Kamila [148]

Answer:

955 in quinary system = 12310

955 in binary system = 1110111011

Step-by-step explanation:

Quinary system is in base 5, that is, 0,1,2,3,4

955 in quinary system

955 ÷ 5 = 191 remainder 0

191 ÷ 5 = 38 remainder 1

38 ÷ 5 = 7 remainder 3

7 ÷ 5 = 1 remainder 2

1 ÷ 5 = 0 remainder 1

In reverse order of the remainder

= 12310

Binary system = 0,1

955 in binary system

955 ÷ 2 = 477 remainder 1

477 ÷ 2 = 238 remainder 1

238 ÷ 2 = 119 remainder 0

119 ÷ 2 = 59 remainder 1

59 ÷ 2 = 29 remainder 1

29 ÷ 2 = 14 remainder 1

14 ÷ 2= 7 remainder 0

7 ÷ 2 = 3 remainder 1

3 ÷ 2 = 1 remainder 1

1 ÷ 2 = 0 remainder 1

In reverse order of the remainder

1110111011

5 0

3 years ago

6/40 with the remainder? <br> (sorry if its 40/6 instead)

mrs_skeptik [129]

Answer:

6.66666666667

Step-by-step explanation:

40 divided by 6 is equal to 6 with a remainder of 4: 40 / 6 = 6 R. 4.

6 0

3 years ago

Forces of 9 lbs and 13 lbs act at a 38º angle to each other. Find the magnitude of the resultant force and the angle that the re

fiasKO [112]

Answer: $R=20.84\ lb\quad 22.57^{\circ},15.43^{\circ}$

Step-by-step explanation:

Given

Two forces of 9 and 13 lbs acts $38^{\circ}$ angle to each other

The resultant of the two forces is given by

$\Rightarrow R=\sqrt{a^2+b^2+2ab\cos \theta}$

Insert the values

$\Rightarrow R=\sqrt{9^2+13^2+2(9)(13)\cos 38^{\circ}}\\\Rightarrow R=\sqrt{81+169+184.394}\\\Rightarrow R=\sqrt{434.394}\\\Rightarrow R=20.84\ lb$

Resultant makes an angle of

$\Rightarrow \alpha=\tan^{-1}\left( \dfrac{b\sin \theta}{a+b\cos \theta}\right)\\\\\text{Considering 9 lb force along the x-axis}\\\\\Rightarrow \alpha =\tan^{-1}\left( \dfrac{13\sin 38^{\circ}}{9+13\cos 38^{\circ}}\right)\\\\\Rightarrow \alpha =\tan^{-1}(\dfrac{8}{19.244})\\\\\Rightarrow \alpha=22.57^{\circ}$