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o-na [289]
3 years ago
8

What is 0.688 as a percent

Mathematics
2 answers:
Lemur [1.5K]3 years ago
7 0
It's 68.8%. You move the decment right two. 
MissTica3 years ago
4 0
68.8% is your answer

when changing a decimal to a percentage, move the decimal point twice over

hope this helps
You might be interested in
The number of gallons of water in a large tank at time (t) minutes is given by the function W(t)=160,000-8,000t+t². Find the ave
sergij07 [2.7K]

The number of gallons of water in the tank at t=10 is

... W(10) = 160,000 -10(8000 -10) = 80100

The number of gallons of water in the tank at t=10.5 is

... W(10.5) = 160,000 -10.5(8000 -10.5) = 76110.25

The rate of change over the interval is

... (W(10.5) - W(10))/(10.5 - 10) = (76110.25 - 80100)/(0.5) = -7979.5

The average rate of change in the number of gallons of water in the tank over the interval is -7979.5 gal/min.

The sign is negative, so the amount of water is decreasing.

3 0
3 years ago
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 =

\^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})

...(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:

brainly.com/question/4608972

#SPJ4

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
1 year ago
The polygons in each pair are similar. find the missing side length​
balandron [24]

Answer:

Missing length = 20

Step-by-step explanation:

Ratio is 16/20=4/5

25×4/5=20

8 0
2 years ago
The sum of two and the quotient of a number I and five.
Alona [7]

Answer:

2+n/5

Step-by-step explanation:

sum=plus

quotient=divide

number=n

4 0
2 years ago
A jar has 20 marbles: 3 green, 12 blue, 5 red.
nadezda [96]

Probability is defined as the <u>likelihood or the certainty</u> that an event is going to<u> occur or happen.</u>

The probability of randomly choosing a red and then a green marble is 3/76.

The total number of marbles = 20

The number of green marbles= 3

The number of blue marbles = 12

The number of red marbles = 5

<u>The probability of choosing a red marble</u> = Number of red marbles / Total number of marbles

= 5/20

<u>In simplest fraction form</u> = \frac{1}{4}

We are told in the question that you keep the red marble you choose, So this means the <u>total number of marbles</u> left reduces to 19

<u>The probability of choosing a green marble is</u> =  Number of green marbles / New total number of marbles

= 3/19

Therefore, <u><em>the probability of randomly choosing a red and then a green marble is </em></u>

P (Red) x P(Green)

= 1/4  x 3/19

= 3/76

To learn more, visit the link below:

brainly.com/question/22563776

4 0
2 years ago
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