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Svetradugi [14.3K]

3 years ago

12

You are relieved from your position as a machine operator at the plant at 11:25 for a 50-minute lunch break. When must you begin

working again.

1 answer:

anastassius [24]3 years ago

5 0

Answer:

12:15

Step-by-step explanation:

You might be interested in

Insert parentheses in expression 2 so that it has a value of 4. Then show why your expression has a value of 4.

Natasha_Volkova [10]

(2^3 +4) = 12

9-6=3

(2^3 +4) /(9-6)
12 /3
4

8 0

3 years ago

Help me pls and dont guess

BARSIC [14]

Answer:

C

Step-by-step explanation:

3 1/2 times -(4/5)

First multiply 3 1/2 and -(4/5)=2 4/5

6 0

3 years ago

Read 2 more answers

Amelia wants to paint three walls in her family room. Two walls are 26 feet long by 9 feet wide. The other wall is 18 feet long

sattari [20]

Given:
Wall 1 & 2 : 26 ft by 9 ft
Wall 3 : 18 ft by 9 ft

Area 1 : 26 ft x 9 ft = 234 ft²
Area 2 : 26 ft x 9 ft = 234 ft²
Area 3 : 18 ft x 9 ft = 162 ft²

Total Area : 234 ft² + 234 ft² + 162 ft² = 630 ft²

1 Gallon = 250 ft²

630 ft² ÷ 250 ft² per gallon = 2.52 gallons

6 0

3 years ago

The volume of the prism is 900 cubic centimeters. The height of the prism is 15 centimeters and the height of the triangular bas

NeTakaya

900=(1/2)*b*10*15
1800=150b
b=12
I hope you can figure out FC with this info.

7 0

3 years ago

Read 2 more answers

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