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katovenus [111]
3 years ago
11

Peterson Packaging Inc. does not currently pay dividends. The company will start with a ​$0.70 dividend at the end of year three

and grow it by 10​% for each of the next six years. After six years of​ growth, it will fix its dividend at ​$1.36 forever. If you want a 15​% return on this​ stock, what should you pay today given this future dividend​ stream?
Mathematics
1 answer:
charle [14.2K]3 years ago
5 0
$1.40 because I think so
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The measure of central angle YCZ is 80 degrees
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The answer is 36 units2

Step-by-step explanation:

I took the test so I’m right

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Document 3 The Fundamental Orders of Connecticut (1639)
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3 years ago
Pls help me calculate the total area of compound shapes?!?!?
Elza [17]

Answer:

Left: 42ft^2

Right: 21ft^2

Step-by-step explanation:

Its the same question as last time.

Left: (3*10) + (2*6) = 42

Right: (3*5) + (2*3) = 21

4 0
1 year ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.
sweet-ann [11.9K]

Answer:

2.794

Step-by-step explanation:

Recall that if G(x,y) is a parametrization of the surface S and F and G are smooth enough then  

\bf \displaystyle\iint_{S}FdS=\displaystyle\iint_{R}F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})dxdy

F can be written as

F(x,y,z) = (xy, yz, zx)

and S has a straightforward parametrization as

\bf G(x,y) = (x, y, 3-x^2-y^2)

with 0≤ x≤1 and  0≤ y≤1

So

\bf \displaystyle\frac{\partial G}{\partial x}= (1,0,-2x)\\\\\displaystyle\frac{\partial G}{\partial y}= (0,1,-2y)\\\\\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y}=(2x,2y,1)

we also have

\bf F(G(x,y))=F(x, y, 3-x^2-y^2)=(xy,y(3-x^2-y^2),x(3-x^2-y^2))=\\\\=(xy,3y-x^2y-y^3,3x-x^3-xy^2)

and so

\bf F(G(x,y))\cdot(\displaystyle\frac{\partial G}{\partial x}\times\displaystyle\frac{\partial G}{\partial y})=(xy,3y-x^2y-y^3,3x-x^3-xy^2)\cdot(2x,2y,1)=\\\\=2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2

we just have then to compute a double integral of a polynomial on the unit square 0≤ x≤1 and  0≤ y≤1

\bf \displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}(2x^2y+6y^2-2x^2y^2-2y^4+3x-x^3-xy^2)dxdy=\\\\=2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}ydy+6\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}x^2dx\displaystyle\int_{0}^{1}y^2dy-2\displaystyle\int_{0}^{1}dx\displaystyle\int_{0}^{1}y^4dy+\\\\+3\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}x^3dx\displaystyle\int_{0}^{1}dy-\displaystyle\int_{0}^{1}xdx\displaystyle\int_{0}^{1}y^2dy

=1/3+2-2/9-2/5+3/2-1/4-1/6 = 2.794

5 0
3 years ago
Help with #2, thank you :)
marishachu [46]

Answer:

2^6

Step-by-step explanation:

2^2 + 2^4 = 2^6

4 0
2 years ago
Read 2 more answers
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