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Bezzdna [24]
3 years ago
6

The product of two consecutive negative integers is 600. What is the value of the lesser integer?

Mathematics
2 answers:
lapo4ka [179]3 years ago
8 0
X ·( x + 1 ) = 600
x² + x = 600
x² + x - 600 = 0
x_{12} = \frac{-1+/- \sqrt{1+2400} }{2}= \frac{-1+/- \sqrt{2401} }{2}= \\  =\frac{-1+/-50}{2}
x = ( - 1 - 49 ) / 2 = - 50 : 2 = - 25
x = - 25,  x + 1 = - 24
( - 25 ) · ( - 24 ) = 600
Answer:
C )   - 25
Likurg_2 [28]3 years ago
5 0

x ·( x + 1 ) = 600

x² + x = 600

x² + x - 600 = 0

x_{12} = \frac{-1+/- \sqrt{1+2400} }{2}= \frac{-1+/- \sqrt{2401} }{2}= \\  =\frac{-1+/-50}{2}  

x = ( - 1 - 49 ) / 2 = - 50 : 2 = - 25

x = - 25,  x + 1 = - 24

( - 25 ) · ( - 24 ) = 600

Answer:

C )   - 25


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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

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