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artcher [175]
3 years ago
5

A garden measuring 12 meters by 6 meters is going to have a walkway constructed all around the perimeter, increasing the total a

rea to 160 square meters. What will be the width of the pathway? (The pathway will be the same width around the entire garden).
Mathematics
2 answers:
damaskus [11]3 years ago
4 0

Answer:

x=2

Step-by-step explanation:

Original width = 6

New width 6+x+x

Orignal length 12

New length  12+x+x

A = l*w

160 = ( 6+2x) ( 12+2x)

Factor

160 = 2( 3+x) 2(6+x)

Divide each side by 4

40 = (3+x) (6+x)

FOIL

40 = 18+ 6x+3x+ x^2

40 = 18 +9x+x^2

Subtract 40 from each side

0 = x^2 +9x -22

Factor

0 = (x +11) (x-2)

Using the zero product property

x +11 =0   x-2 =0

x= -11   x=2

Since we cannot have a negative  sidewalk

x =2

Paraphin [41]3 years ago
3 0

Answer:

2

Step-by-step explanation:

Original width = 6

New width = 6 + x + x = 6 + 2x

Orignal length = 12

New length = 12 + x + x = 12 + 2x

A = l * w

160 = (6 + 2x)(12 + 2x)

160 = 2(3+x) * 2(6+x)

160 = 4 * (3 + x)(6 + x)

160/4 = (3 + x)(6 + x)

40 = 18 + 6x + 3x + x^2

40 = 18 + 9x + x^2

x^2 + 9x - 22 = 0

= x^2 + 11x - 2x - 22 = 0

= x(x + 11) - 2(x + 11) = 0    

= (x + 11) (x - 2) = 0

x = - 11, 2

Since we cannot have a negative width because it's a dimension,

x = 2 is right

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\bold{\huge{\underline{ Solution\: 2 }}}

  • The name of the circle is <u>IJKL</u>
  • The name of central Angle
  • \sf{\underline{= {\angle} IHJ }}
  • The name of the secant = <u>KL</u>

{\bigstar}Circle :- It is a round figures, which has no end points.

{\bigstar} Central angle :- It the angle ,which lie in the middle of the circle.

  • Here, H is the mid point of the circle .So, Angle IHJ is the central angle.

{\bigstar} Secant :- A line drawn in circle in such a way that it intersect the circle from two distinct points. Then, It is called secant .

  • Here, KL is the line which acts as a secant because it intersect the circle from point K and point L.

\bold{\huge{\underline{ Solution\: 3 }}}

  • Semicircle = FGH and FIH
  • Minor Arc :- FG or GH
  • Major Arc :- FHG or GFH

{\bigstar} Semicircle :- Semicircle is nothing but the half of the circle.

<u>Here</u><u>, </u><u>In </u><u>second </u><u>diagram </u>

  • FGH and FIH are acting as two semicircles which simulatineously forming one circle.

{\bigstar} Minor Arc :- Minor Arc is nothing but the smaller arc of the circle and it is always smaller than the half of the circle that is semicircle.

<u>Here</u><u>, </u>

  • FG or GH are the minor arc of the given circle.

{\bigstar} Major Arc :- Major Arc is nothing but the largest arc of the circle and it is always larger than the half of the circle that is semicircle.

<u>Here</u><u>, </u>

  • FHG or GFH is the major arc as it is larger than the semicircle of the given circle .
7 0
2 years ago
Can someone please help me with #2?​
Norma-Jean [14]

For problem 2, you are correct in stating that a curve forms. Specifically, if we were to trace along the outer edge of the shape, then we'd form a <u>parabola</u> that has been tilted 45 degrees compared to the more familiar form that students are taught (where the axis of symmetry is vertical).

For more information, search out "Tangent method for parabolas". As the name implies, the tangent method draws out the tangents of the parabola which helps form the parabola itself.

Everything else on your paper is correct. You have problem 1 correct, and the table is filled out perfectly. Nice work.

4 0
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