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3 years ago

What are the counting numbers in base 6

1 answer:

8 0

In the base 6 number system, we would only use numerals from 0 to 5. The idea of base 6 is just like the normal base 10 system, however instead of using the numerals from 0 to 9, we use numerals from 0 to 5. And instead of having a ones digit, a tens digit, a hundreds digit and so on, we use a ones digit, a sixes digit, a thirty-sixes digit, and so on. Therefore in base 6, the number 321 means 1 one plus 2 sixes plus 3 thirty-sixes, or 121. So to count until 15 for example, we would need:

Base 6 Base 10

1 1

2 2

3 3

4 4

5 5

10 (1 six plus 0 ones) 6

11 (1 six plus 1 one) 7

12 (1 six plus 2 ones) 8

13 9

14 10

15 11

20 (2 sixes plus 0 ones) 12

21 (2 sixes plus 1 one) 13

22 14

23 15

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Let and be differentiable vector fields and let a and b be arbitrary real constants. Verify the following identities.

elena-14-01-66 [18.8K]

The given identities are verified by using operations of the del operator such as divergence and curl of the given vectors.

<h3>What are the divergence and curl of a vector field?</h3>

The del operator is used for finding the divergence and the curl of a vector field.

The del operator is given by

$\nabla=\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}$

Consider a vector field $F=x\^i+y\^j+z\^k$

Then the divergence of the vector F is,

div F = $\nabla.F$ = $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(x\^i+y\^j+z\^k)$

and the curl of the vector F is,

curl F = $\nabla\times F$ = $\^i(\frac{\partial Fz}{\partial y}- \frac{\partial Fy}{\partial z})+\^j(\frac{\partial Fx}{\partial z}-\frac{\partial Fz}{\partial x})+\^k(\frac{\partial Fy}{\partial x}-\frac{\partial Fx}{\partial y})$

<h3>Calculation:</h3>

The given vector fields are:

$F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$

1) Verifying the identity: $\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

Consider L.H.S

⇒ $\nabla.(aF1+bF2)$

⇒ $\nabla.(a(M\^i + N\^j + P\^k) + b(Q\^i + R\^j + S\^k))$

⇒ $\nabla.((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the dot product between these two vectors,

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(1)

Consider R.H.S

⇒ $a\nabla.F1+b\nabla.F2$

So,

$\nabla.F1=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(M\^i + N\^j + P\^k)$

⇒ $\nabla.F1=\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z}$

$\nabla.F2=(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z}).(Q\^i + R\^j + S\^k)$

⇒ $\nabla.F1=\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z}$

Then,

$a\nabla.F1+b\nabla.F2=a(\frac{\partial M}{\partial x}+\frac{\partial N}{\partial y}+\frac{\partial P}{\partial z})+b(\frac{\partial Q}{\partial x}+\frac{\partial R}{\partial y}+\frac{\partial S}{\partial z})$

⇒ $\frac{\partial (aM+bQ)}{\partial x}+ \frac{\partial (aN+bR)}{\partial y}+\frac{\partial (aP+bS)}{\partial z}$ ...(2)

From (1) and (2),

$\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2$

2) Verifying the identity: $\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Consider L.H.S

⇒ $\nabla\times(aF1+bF2)$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times(a(M\^i+N\^j+P\^k)+b(Q\^i+R\^j+S\^k))$

⇒ $(\^i\frac{\partial}{\partial x}+\^j \frac{\partial}{\partial y}+\^k\frac{\partial}{\partial z})\times ((aM+bQ)\^i+(aN+bR)\^j+(aP+bS)\^k)$

Applying the cross product,

$\^i(\^k\frac{\partial (aP+bS)}{\partial y}- \^j\frac{\partial (aN+bR)}{\partial z})+\^j(\^i\frac{\partial (aM+bQ)}{\partial z}-\^k\frac{\partial (aP+bS)}{\partial x})+\^k(\^j\frac{\partial (aN+bR)}{\partial x}-\^i\frac{\partial (aM+bQ)}{\partial y})$ ...(3)

Consider R.H.S,

⇒ $a\nabla\times F1+b\nabla\times F2$

So,

$a\nabla\times F1=a(\nabla\times (M\^i+N\^j+P\^k))$

⇒ $\^i(\frac{\partial aP\^k}{\partial y}- \frac{\partial aN\^j}{\partial z})+\^j(\frac{\partial aM\^i}{\partial z}-\frac{\partial aP\^k}{\partial x})+\^k(\frac{\partial aN\^j}{\partial x}-\frac{\partial aM\^i}{\partial y})$

$a\nabla\times F2=b(\nabla\times (Q\^i+R\^j+S\^k))$

⇒ $\^i(\frac{\partial bS\^k}{\partial y}- \frac{\partial bR\^j}{\partial z})+\^j(\frac{\partial bQ\^i}{\partial z}-\frac{\partial bS\^k}{\partial x})+\^k(\frac{\partial bR\^j}{\partial x}-\frac{\partial bQ\^i}{\partial y})$

Then,

$a\nabla\times F1+b\nabla\times F2$ =

...(4)

Thus, from (3) and (4),

$\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

Learn more about divergence and curl of a vector field here:

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Disclaimer: The given question on the portal is incomplete.

Question: Let $F1 = M\^i + N\^j + P\^k$ and $F2 = Q\^i + R\^j + S\^k$ be differential vector fields and let a and b arbitrary real constants. Verify the following identities.

$1)\nabla.(aF1+bF2)=a\nabla.F1+b\nabla.F2\\2)\nabla\times(aF1+bF2)=a\nabla\times F1+b\nabla\times F2$

8 0

2 years ago

Price of 1kg sugar is Rs. 100 calculate the price of 150g sugar?

Wewaii [24]

Answer:Rs15

Step-by-step explanation:

150g=150/1000=0.15kg

1kg cost Rs100

0.15kg cost =100 x 0.15=15

150g cost Rs15

3 0

3 years ago

-1. The table shows the annual consumption of cheese per person in the United States for selected years in the 20th century. Let

olga nikolaevna [1]

<span>Using processing software (Excel) or even a decent scientific calculator. You input the values and generate the best fit cubic equation.
For number 1, the equation is
y = 8x10</span>⁻⁵ x³ - 0.0097 x² + 0.374 x + 1.083
where x is the number of years since 1900
y is the pounds cheese consumed

For number 2, the equation is
y = -3x10⁻⁵ x³ + 0.0028 x² + 0.2155 x + 1.7736

For number 3
P(-1) = 18

5 0

3 years ago

Having a hard time with this can someone answer pls

larisa [96]

The values of angles are 140 and 20

• In plane geometry, a figure which is formed by joining of two lines that share a common point is called as the angle. The two lines or rays are called as the sides of the angle and the common point is called the vertex.

• In geometry there are types of angles such as complementary angles, supplementary angles, acute angle, obtuse angles. Complementary angles are the angles whose sum is equal to 90. Supplementary angles are the angles whose sum is equal to 180.

According to the question

We are given that one angle is -7x and the other angle is -2x

Using formula of supplementary angles

-7x + (-2x) = 180

-7x -2x = 180

-9x = 180

-x = 20

x = -20

The value of x is -20

The value of angle – 7x = -7(-20) = 140

The value of angle -2x = -2(-20) = 40

Learn more about angles here

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3 0

2 years ago

Verda used a sensor to measure the speed of a moving car at different times. At

Kaylis [27]

Answer:

the answer is c because if you divided 4848484 by 37 =234

Step-by-step explanation:

393-=3\3

3 0

3 years ago