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3 years ago

If licorice cost $6.59 a pound how much would it cost to buy a quarter pound of licorice

Mathematics

2 answers:

schepotkina [342]3 years ago

8 0

licorice cost $6.59 a pound

how much would it cost to buy a quarter pound of licorice

$6.59 / 4 = 1.6475

a quarter pound of licorice

$1.65

RideAnS [48]3 years ago

7 0

Answer:

$1.6475

Step-by-step explanation:

divide 6.59 by 4

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What is: 13 to the 3rd power 2/5 to the second power And 0.9 to the second power

Anna007 [38]

1. 13*13*13 =2197
2. 2/5*2/5 = 4/25
3. 0.9*0.9 = 0.81

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3 years ago

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atroni [7]

Answer:

Step-by-step explanation:

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2 years ago

Enter an inequality that represents the graph in the box.

ANTONII [103]

The inequality that represents the graph in the box is

$y>-3x+6$

Given :

The graph of the linear inequality

Lets pick two points from the graph to frame the linear equation

Slope intercept form of the equation is

$y=mx+b$

Where m is the slope and b is the y intercept

y intercept is (0,6) from the graph

so b=6

Now we find out slope m using formula

Pick two points (0,6) and (2,0)

$slope = \frac{y_2-y_1}{x_2-x_1} \\m=\frac{0-6}{2-0} =-3$

m=-3

So the equation is $y=-3x+6$

Now we check the inequality sign by testing any point on shaded area

lets pick (4,0)

lets plug in 4 for x and 0 for y

$y=-3x+6\\0=-3(4)+6\\\\0=-6\\0>-6$

The inequality for the given graph is

$y>-3x+6$

we cannot use >= sign because we have dotted lines

Learn more : brainly.com/question/17203372

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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

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

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

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3 years ago

Solve for x in the equation x2 -8x+41=0. X=-4+ 37i X = -4 - 5i X = 4+ 37i

Digiron [165]

Answer:

not sure if this is a right answer. but hope this helps

Step-by-step explanation:

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3 years ago