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Misha Larkins [42]
3 years ago
8

The definition of parallel lines requires the undefined terms line and plane, while the definition of perpendicular lines requir

es the undefined terms of line and point. What characteristics of these geometric figures create the different requirements?
Mathematics
1 answer:
Ludmilka [50]3 years ago
8 0

Answer:

Parallel lines never intersect, but they must be in the same plane. The definition does not require the undefined term point, but it does require plane. Because they intersect, perpendicular lines must be coplanar; consequently, plane is not required in the definition.

Step-by-step explanation:

Its correct trust me.

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Find percent notation<br> 0.488
In-s [12.5K]

Answer:

48.8%

Step-by-step explanation:

Just multiply this 0.488 by 100%, obtaining 48.8%.

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Which expression is a factor of 9x^2-4x+1 ( ps ^2 is an exponent)
erastovalidia [21]

Answer:

The discriminant b2−4ac < 0, the equation has no real number solutions, it has complex solutions

Step-by-step explanation:

9x² - 4x + 1 = 0     ax² + bx + c = 0

The discriminant: b² - 4ac = (-4)² - 4*9*1 = 16 - 36 = -20

b2−4ac < 0, the equation has no real number solutions, it has complex solutions

3 0
3 years ago
If p is true and q is true, then ^p — ^ ~ q is true.
Ede4ka [16]

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5 0
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Read 2 more answers
Reciprocate 2 5/9 <br>plz and thx​
ZanzabumX [31]

Answer: 9/23

Step-by-step explanation:

1- Convert the mixed fraction to an improper fraction.  9 x 2 + 5 = 23 and keep the original denominator.  23/9

2- Now “flip” the fraction to its reciprocal. 23/9 becomes 9/23

3- Check to see if it can be reduced.  It cannot so there’s your answer.  9/23

6 0
3 years ago
Given x^2 + y^2 = 36 and xy = -10 find x + y.
9966 [12]
This looks fun

ok so
we will use subsitution
xy=-10
divide both sides by x or y (I will choose y)
x=-10/y

sub -10/y fo x

(-10/y)^2+y^2=36
100/(y^2)+y^2=36
times both sides by y^2
100+y^4=36y^2
minus 36y^2 from both sides
y^4-36y^2+100=0
(y^2)^2-36(y^2)+100=0
quadratic formula
for
ax^2+bx+c=0
x=\frac{-b+/- \sqrt{b^2-4ac} }{2a}
x=\frac{-(-36)+/- \sqrt{(-36)^2-4(1)(100)} }{2(1)}
x=18+/- 4\sqrt{14}

sub
y=-10/x
y=\frac{-10}{18+/- 4\sqrt{14}}

so x+y=18+/- 4\sqrt{14}+\frac{-10}{18+/- 4\sqrt{14}}
multiply first number by \frac{18+/- 4\sqrt{14}}{18+/- 4\sqrt{14}} and add them

x+y=\frac{(18+/- 4\sqrt{14})(18+/- 4\sqrt{14})-10}{18+/- 4\sqrt{14}}
or
x+y=\frac{81+22 \sqrt{14} }{5} or \frac{81-22 \sqrt{14} }{5}




8 0
3 years ago
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