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igor_vitrenko [27]
4 years ago
7

If 22000 is invested at 12% interest compounded quarterly, find the interest earned in 12 years.

Mathematics
1 answer:
ICE Princess25 [194]4 years ago
7 0

Answer:

Interest earned in 12 years = 57178

Step-by-step explanation:

A=P(1+rn)n⋅t

A = total amount

P = principal or amount of money deposited,

r = annual interest rate

n = number of times compounded per year

t = time in years

then

P=22000 , r=12% , n=12 and t=12 years

putting these value

A=22000(1+0.12/12) 12⋅8

        = 22000(2.5992729255593856)

        = 57178

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Compute the line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the l
SIZIF [17.4K]

Answer:

\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}

Step-by-step explanation:

The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.

Let  

C_1,C_2  

be the two paths.

Recall that if we parametrize a path C as (r_1(t),r_2(t),r_3(t)) with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{a}^{b}f(r_1,r_2,r_3)\sqrt{(r'_1)^2+(r'_2)^2+(r'_3)^2}dt

Given any two points P, Q we can parametrize the line segment from P to Q as

r(t) = tQ + (1-t)P with 0≤ t≤ 1

The parametrization of the line segment from (1,1,1) to (2,2,2) is

r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)

r'(t) = (1,1,1)

and  

\displaystyle\int_{C_1}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(1+t,1+t,1+t)\sqrt{3}dt=\\\\=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)(1+t)^2dt=\sqrt{3}\displaystyle\int_{0}^{1}(1+t)^3dt=\displaystyle\frac{15\sqrt{3}}{4}

The parametrization of the line segment from (2,2,2) to  

(-9,6,5) is

r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)  

r'(t) = (-11,4,3)

and  

\displaystyle\int_{C_2}f(x,y,z)ds=\displaystyle\int_{0}^{1}f(2-11t,2+4t,2+3t)\sqrt{146}dt=\\\\=\sqrt{146}\displaystyle\int_{0}^{1}(2-11t)(2+4t)^2dt=-90\sqrt{146}

Hence

\displaystyle\int_{C}f(x,y,z)ds=\displaystyle\int_{C_1}f(x,y,z)ds+\displaystyle\int_{C_2}f(x,y,z)ds=\\\\=\boxed{\displaystyle\frac{15\sqrt{3}}{4}-90\sqrt{146}}

8 0
3 years ago
What is the proper seperation of factors for this equation ∛24n² × ∛36n²
Gwar [14]

Answer:

{ \tt{ \sqrt[3]{24 {n}^{2}  }  \times  \sqrt[3]{36 {n}^{2} } }} \\  = { \tt{(  {n}^{ \frac{2}{3} })( \sqrt[3]{864})  }}

8 0
3 years ago
(Use the Distributive Property to find an equivalent expression)<br> 2(3x-7y) = ?
IRINA_888 [86]

Answer:

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Step-by-step explanation:

It's the same thing but backwards

I hope this helps

6 0
3 years ago
HELP!!!!
Nonamiya [84]
Your answer is E. $25.

First let under 12 = u, over 12 = o, and adults = a.
We can now write the equations:

2u + 3a + 3o = 174
4u + 2a = 122
a + o = 46

Because we know that a + o = 46, and 3a + 3o is in the first equation, we can multiply 46 by 3 to get what 3a + 3o equals. This makes 138.
Now we can substitute 138 into the first equation to get 2u + 138 = 174
2u = 36
u = 18

Now that we know what u equals, we can substitute it in to the second equation to get:
4(18) + 2a = 122
72 + 2a = 122
2a = 50
a = $25

I hope this helps! Let me know if you have any questions :)
8 0
3 years ago
Find the slope of a line
Dvinal [7]

Answer:

1 is the slope.

Step-by-step explanation:

I did this math question on edge.

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