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wel
4 years ago
14

Spencer has a rectangular garden. The length and width are both odd numbers and the perimeter is 16 feet. What can be the area o

f his garden? Find all possible answers.
Mathematics
1 answer:
vovikov84 [41]4 years ago
5 0

Area of rectangle = length x width

Perimeter of rectangle = 2 x (length + width)

Now we know that perimeter of rectangle = 16

2(length + width) = 16

Length + width = 16/2

Length + width = 8

Now the value of length and width is that if we add both values we have the answer 8.

So, here are some possible values of length and width and also that length and width are odd numbers;

Odd numbers are = 1,3,5,7,….

7 + 1 = 8

5 + 3 = 8

So, either length is 7 and with is 1 or length is 5 and width is 3

So area of rectangle = 7 x 1 = 7 or 5 x 3 = 15

Thus, the possible values of area of rectangle are 7 ft² and 15 ft².

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The slope of EF¯¯¯¯¯ is −52.
9966 [12]

Based on the above, the segments that are perpendicular to EF are LM and NP.

<h3>Why is the segment are LM and NP perpendicular  to EF ?</h3>

Note that when two lines are perpendicular, we can say that;

M1 * M2 = -1 As M1 and M2 are known to be the slopes of the lines.

Therefore, when the the slope of EF is said to be −5/2, then  one can say that the slope of  the segment that is said to be  perpendicular to EF will have to be equal to m1*m2=-1, m2=-1/m1, m2=-1/(-5/2) or m2=2/5.

Scenario one:

JK , if J is at (3, −2) and K is at (5, −7)

To find the slope JK, then

m=(y2-y1)/x2-x1)

m=(-7+2)/(5-3)

m=-5/2

-5/2 is not equal to 2/5

Therefore,  JK is not perpendicular to EF

Scenario 2

Find GH , when G is at (6, 7) and H is at (4, 12)

To find the slope GH

m=(y2-y1)/x2-x1)

m=(12-7)/(4-6)

m=5/-2

m=-5/2

Since -5/2 is not equal to 2/5 then GH is not perpendicular to EF

Scenario 3:

Find LM , If L is at (1, 9) and M is at (6, 11)

To find the slope LM, then

m=(y2-y1)/x2-x1)

m=(11-9)/(6-1)

m=2/5

Since 2/5 is equal to 2/5

Then LM is perpendicular to EF

Scenario 4:

Find NP , if N is at (−3, 4) and P is at (−8, 2)

To find the slope NP, then

m=(y2-y1)/x2-x1)

m=(2-4)/(-8+3)

m=-2/-5

m=2/5

Since 2/5 is equal to 2/5.

Therefore,  NP is perpendicular to EF

Based on the above calculations, the segments that are perpendicular to EF are LM and NP.

See correct format of question written below

The slope of EF is −5/2 .

Which segments are perpendicular to EF?

Select all the right answers please

1. JK , where J is at (3, −2) and K is at (5, −7)

2. GH , where G is at (6, 7) and H is at (4, 12)

3. LM , where L is at (1, 9) and M is at (6, 11)

4. NP , where N is at (−3, 4) and P is at (−8, 2)

Learn more about segments from

brainly.com/question/10565562

#SPJ1

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Answer:

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Answer:

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