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Nataliya [291]
3 years ago
11

Mike drank too much at a party las night this morning he can’t remember the previous nights events which term describes mikes ex

perience
Mathematics
1 answer:
bulgar [2K]3 years ago
3 0
Well his experience is that he had some drink (liquor we know this) the substance he had was not good to his brain. he proboarly blacked out.

 WARNING: DO NOT PUT DOWN THE LIQUOR PART
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Find the equation of the linear function represented by the table below in the slope-intercept form.
MAXImum [283]

Answer:

y=x-32

Step-by-step explanation:

y=mx+b where m is the slope and b is the y-intercept.

slope=change in y/change in x

=5/5

y=5x+b --> choose a random point

(7,3) ---> 3=5(7)+b=35+b

b= -32

y=x-32

6 0
2 years ago
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Please help im so confused lol
BARSIC [14]

Answer:

1. Percent

2. Number

Step-by-step explanation:

6 0
3 years ago
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What is the area of this.
Ratling [72]
8x16=? 6x12=? Then add both answers ?+?=?
8 0
3 years 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
What is the solution, if any, to the inequality |3x| ≥0?
Yanka [14]

Answer:

the answer is either c or b

4 0
2 years ago
Read 2 more answers
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