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3 years ago

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Jonathan and Sarah purchased a beach house for $10,000,000. The home appreciates at a rate of 3% per year. The appreciation of t

he house is shown in the graph. The home is currently worth about $13,000,000. Jonathan and Sarah have owned the home for about ______ years. how many years

2 answers:

mars1129 [50]3 years ago

5 0

13000000=10000000 (1+0.03)^t
Solve for t
T=log(13,000,000÷10,000,000)÷log(1+0.03)
t=8.9 years round 9 years

ValentinkaMS [17]3 years ago

4 0

Current worth = 10,000,000 * (1.03)^years
We must determine the years when current worth = 13,000,000
13,000,000 / 10,000,000 = 1.03^years
1.3 = 1.03^years
Taking logs of both sides:
log (1.3) = years*log(1.03)
log(1.3) / log(1.03) = years
0.11394335231 / 0.012837224705 = years
years = <span> <span> <span> 8.8760113598 </span> </span> </span>
Roughly about 8.9 years

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Not sure need help

Greeley [361]

29/90 use the hint:)

6 0

3 years ago

3 years ago a father was 3 times as old as his son, in five years time he will be twice as old as his son, what will be the sum

ivolga24 [154]

Answer: 46 years

Step-by-step explanation:

Let the father's age be x and the son's age be y, then 3 years ago:

Father = x - 3

son = y - 3

Then , from the first statement :

x - 3 = 3 ( y - 3 )

x - 3 = 3y - 9

x = 3y - 9 + 3

x = 3y - 6 .......................................... equation 1

In five years time

father = x + 5

son = y + 5

Then , from the second statement

x + 5 = 2 ( y + 5 )

x + 5 = 2y + 10

x = 2y + 10 - 5

x = 2y + 5 ........................ equation 2

Equating equation 1 and 2 , we have

3y -6 = 2y + 5

add 6 to both sides

3y = 2y + 5 + 6

subtract 2y from both sides

3y - 2y = 11

y = 11

substitute y = 11 into equation 1 to find the value of x

x = 3y - 6

x = 3(11) - 6

x = 33 - 6

x = 27

This means that the father is presently 27 years and the son is presently 11 years.

In four years time

father = 27 + 4 = 31

son = 11 + 4 = 15

sum of their ages in four years time will be

31 + 15 = 46 years

5 0

3 years ago

Read 2 more answers

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

2 years ago

How do you write 0.326 and 0.924 in word form?

Sergeeva-Olga [200]

three hundred twenty-six thousandths = .326
nine hundred twenty-four thousandths = .924

Hope this helped!! :D

4 0

3 years ago

Read 2 more answers

How do you write 19,830 in scientific notation? _____× 10^_____

pshichka [43]

Answer:

1.983(10⁴)

Step-by-step explanation:

Since we need to have ones and decimals as proper scientific notation, we have 1.983. Since we need the value of 19830, we need to move the decimal place 4 places to the right, so our exponent is 4.

*I double checked my work this time.

8 0

3 years ago

Read 2 more answers