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almond37 [142]
3 years ago
14

John has $1,042. He wants to spend all of his money on video games that cost $59 each. How many video games can he buy? How much

money will he have left over?
Mathematics
2 answers:
MrRissso [65]3 years ago
7 0

Answer:

17 video games

Step-by-step explanation:

To find how much video games John can buy:

1, 042 divided by 59 = 17. 66 but since you can't buy a part of a video game, it's 17.

To check how much money you have left over:

Multiply 59 by 17 = 1, 003.

1, 003 is how much money he paid for 17 video games.

$1,042 - $1,003 = $39 left.

Hope you understand!


Tcecarenko [31]3 years ago
3 0

Answer:he can buy 17 games with $39 left over


Step-by-step explanation:


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Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S.
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Step-by-step explanation:

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\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)

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\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)

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\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

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