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fenix001 [56]
3 years ago
14

A rectangular garden has length twice as great as its width. A second rectangular garden has the same width as the first garden

and length that is 4 meters greater than the length of the first garden. The second garden has area of 70 square meters.
What is the width of the two gardens?

Enter your answer in the box.


m
Mathematics
1 answer:
stiks02 [169]3 years ago
6 0
Use the information to write the dimensions of each rectangle in terms of w, the width of the 1st one.

1st rectangle;
l = 2w
w = 2

2nd rectangle:
w = w 
l = 2w + 4

If the area of the 2nd rectangle is 70 square meters, you will use the area formula to write an equation that you will solve using the factoring.

A = lw
70 = w(2w + 4)
70 = 2w^2 + 4w
0 = 2w^2 + 4w -70
0 = 2 (w^2 +2w - 35)
0 = 2 (w + 7) (w - 5)

To get zero, the width would need to be -7 or 5.  Because it is a distance, it has to be 5 meters.

The width of both rectangles is 5 meters.

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Y=5/2x-19 and 5x+2y=-8 a perpendicular,parallel or intersecting lines?
dimaraw [331]

Answer:

There intersecting lines

Step-by-step explanation:

5x+2y=-8

-5x            -5x

2y=-8 -5x

/2   /2  /2

y=-4 -5/2x also

y=-5/2x - 4

&

y=5/2x - 19

neither parallel (slope is not the same)

or perpendicular (slope is not reciprocal or opposite signed)

hope i helped

6 0
3 years ago
Which of the following models the Associative Property?
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It’s d because it’s d because it’s d because it’s d
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1. Bob’s age is twice that of his sister. When you add Bob's age to his sister's age, you get 24. How old is Bob and his sister?
zmey [24]

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8 and 16

Step-by-step explanation:

16 + 8 = 24

7 0
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Nadene is incorrect because she took GB/US x 5
3 0
2 years ago
Classify ABC by its sides. Then determine whether it is a right triangle.
m_a_m_a [10]

Answer:

∴Given Δ ABC is not a right-angle triangle

a= AB = √45 = 3√5

b = BC = 12

c = AC = √45 = 3√5

Step-by-step explanation:

Given vertices are A(3,3) and B(6,9)

            AB = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

            AB = \sqrt{(9-3)^{2}+(6-3)^{2}  } = \sqrt{6^{2}+3^{2}  } =\sqrt{45}

Given vertices are  B(6,9) and C( 6,-3)

       B C = \sqrt{x_{2}-x_{1} )^{2} +(y_{2}-y_{1} )^{2}  }

             =  \sqrt{(-3-9)^{2}+(6-6)^{2}  } =\sqrt{12^{2} } = 12

    BC = 12

Given vertices are  A(3,3) and C( 6,-3)

 AC = \sqrt{(6-3)^{2}+(-3-3)^{2}  } = \sqrt{9+36} = \sqrt{45}

AC² = AB²+BC²

45  = 45+144

 45  ≠ 189

∴Given Δ ABC is not a right angle triangle

 

5 0
2 years ago
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