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mihalych1998 [28]
3 years ago
13

Please help me with this Algebra problem: The length of a rectangle is 91 less than six times the width of the rectangle. If the

perimeter of the rectangle is 84 centimeters, what is the length of the rectangle? Write a system of equations for this situation and find its solution.
I have so far: P = 2L + 2W P = 2 (6x-91) + 2W P = 12x -182 + 2W 2 (6x-91) =84 12x -182 =84 12x =266 Please help...I am confused...thank you
Mathematics
1 answer:
horsena [70]3 years ago
5 0
 182 Let x = width then y = 6x - 91. P (perimeter) = 2x + 2y = 84 then

2(x) + 2(6x - 91) = 84 | 2x + 12x - 182 =84 | 14x = 84 + 182 | 14x = 266 | x = 19 cm

y = 6x - 91 | y = 6 (19) - 91 | y = 114 - 91 | y = 23 cm

Check: P = 2x + 2y | P = 2(19) + 2(23) | P = 38 + 46 | P = 84 cm (Checked)
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If  5 = 6 , than 4 =7

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The coordinates of the vertices of / are G(–2, 3), H(–1, 2), and I(–3, 1). If / is reflected across the x-axis to create /, find
valkas [14]

Answer:

A. (–2, –3)

that is the answer

3 0
4 years ago
A cargo plane flew from the US across the Atlantic at 212 mph, and flew back to the US at 232 mph. Given that the first trip too
expeople1 [14]

The return trip was 21.2 hours long.

Step-by-step explanation:

Given,

Speed of plane from US = 212 mph

Speed of plane on return trip = 232 mph

Let,

x be the time taken on return trip.

Time taken from US = x+2

We know that;

Distance = Speed * Time

Distance of plane from US = 212(x+2) = 212x+424

Distance on return trip = 232x

As the distance will be same, therefore,

232x=212x+424\\232x-212x=424\\20x=424

Dividing both sides by 20

\frac{20x}{20}=\frac{424}{20}\\x=12.2\ hours

The return trip was 21.2 hours long.

Keywords: distance, speed

Learn more about speed at:

  • brainly.com/question/12960754
  • brainly.com/question/12967961

#LearnwithBrainly

5 0
4 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

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

5 0
3 years ago
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Gnesinka [82]

Answer:

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Step-by-step explanation:

its supposed to be the same on the bottom, right?

3 0
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