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UkoKoshka [18]
3 years ago
11

How to find the distance between two lines in 3d?

Mathematics
1 answer:
Nastasia [14]3 years ago
7 0
Use graph paper, draw the 3d shape and count the squares as one unit each

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u(x) = x5 – x4 + x2 and v(x) = –x2, which expression is equivalent to (StartFraction u Over v EndFraction) (x)?
EastWind [94]

Answer:

Step-by-step explanation:

Bello,

\dfrac{u(x)}{v(x)}=\dfrac{x^5-x^4+x^2}{-x^2}=-x^3+x^2+1

thanks

5 0
3 years ago
Read 2 more answers
What is the justification for each step in solving the inequality?
alekssr [168]
HELLO  ANNA!!

Given inequality
=> −2(x+1)≥3x+8
=>
- 2x - 2 \geqslant 3x + 8 \\ \\ add \: 2 \: in \: both \: sides \\ \\ - 2x \geqslant 3x + 10 \\ \\ subtract \: with \: - 3x \: in \: both \: sides \: \\ \\ - 5x \geqslant 10 \\

now multiply with - on both sides.

As we know now sign will change
=> 5x >_ -10

Now divide by 5 on both sides
=> x >_ -2
HOPE IT HELPED YOU.
4 0
3 years ago
Read 2 more answers
II
kati45 [8]
The answer will be 0;7 and 2;112.
5 0
3 years ago
I need the answer ASAP
GalinKa [24]

Answer:

the answer is A

Hope it helps

5 0
2 years ago
Write the vector u as a sum of two orthogonal vectors, one of which is the vector projection of u onto v
blagie [28]
U = ( -8 , -8)
v = (-1 , 2 )
<span>the magnitude of vector projection of u onto v =
</span><span>dot product of u and v over the magnitude of v = (u . v )/ ll v ll
</span>
<span>ll v ll = √(-1² + 2²) = √5
</span>
u . v = ( -8 , -8) . ( -1 , 2) = -8*-1+2*-8 = -8 
∴ <span>(u . v )/ ll v ll = -8/√5</span>
∴ the vector projection of u onto v = [(u . v )/ ll v ll] * [<span>v/ ll v ll]
</span>
<span>                          = [-8/√5] * (-1,2)/√5 = ( 8/5 , -16/5 )
</span>
The other orthogonal component = u - ( 8/5 , -16/5 )
                                   = (-8 , -8 ) - <span> ( 8/5 , -16/5 ) = (-48/5 , -24/5 )
</span>
So, u <span>as a sum of two orthogonal vectors will be
</span>
u = ( 8/5 , -16/5 ) + <span>(-48/5 , -24/5 )</span>










3 0
3 years ago
Read 2 more answers
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