Answer:
Both types of lines are built on the same plane although they meet different objectives for which they are built.
Step-by-step explanation:
The difference is that in the case of parallel lines they will never cross and in the case of perpendicular lines if they cross each other, which can give rise to the formation of right angles to each other.
Answer:
0.189
4.525
Step-by-step explanation:
v/1 + 2/v + 4v/3 = 11
we can eliminate denominators if we distribute 3v to each term:
3v(v/1 + 2/v + 4v/3) = 3v(11)
3v² + 6 + 4v² = 33v
7v² - 33v + 6 = 0
factor this trinomial:
there are no factors of 6 that would add to -33 so we can use the quadratic formula to solve: (-b ± √b²-4ac) / 2a
a = 7
b = -33
c = 6
[33 ± √(-33)² - 4(7)(6)] / 2(7)
(33 ± √1089 - 168) / 14
(33 ± √921) / 14
(33 ± 30.35) / 14
(33+30.35) / 14 = 4.525
(33-30.35) / 14 = 0.189
Answer: 1
Step-by-step explanation: you have to set y to zero then solve for x. 6(0) = 0 divide 6x by six on both sides so it should look like this x=1
Answer:
Leorio, Gon, Killua, Kurapika, Hisoka, and Ging
Step-by-step explanation
<h2>Steps:</h2>
So for this, I will be factoring by grouping. Firstly, factor x³ + 2x² and -4x - 8 separately. Make sure that they have the same quantity inside of the parentheses:
![p(x)=x^2(x+2)-4(x+2)](https://tex.z-dn.net/?f=p%28x%29%3Dx%5E2%28x%2B2%29-4%28x%2B2%29)
Now we can rewrite it as:
![p(x)=(x^2-4)(x+2)](https://tex.z-dn.net/?f=p%28x%29%3D%28x%5E2-4%29%28x%2B2%29)
However, we aren't finished factoring yet. The first factor, x² - 4, can be factored further using the difference of squares. The difference of squares goes by the formula here:
. In this case:
![x^2-4=(x+2)(x-2)\\p(x)=(x+2)(x-2)(x+2)](https://tex.z-dn.net/?f=x%5E2-4%3D%28x%2B2%29%28x-2%29%5C%5Cp%28x%29%3D%28x%2B2%29%28x-2%29%28x%2B2%29)
<h2>Answer:</h2>
<u>In short, the answer is
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