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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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Need help with 34,35,36,37
nirvana33 [79]
34. x - 3 = 8x + 4
      -3 - 4 = 8x - x
      -7 = 7x
      -7/7 = x
      -1 = x

y = x - 3
y = -1 - 3
y = -4

solution is (-1,-4)
===================
35. -6 = -x - 2y
      x = -2y + 6

2 + 5x = -2y
2 + 5(-2y + 6) = -2y
2 - 10y + 30 = -2y
32 = -2y + 10y
32 = 8y
32/8 = y
4 = y

2 + 5x = -2y
2 + 5x = -2(4)
2 + 5x = -8
5x = -8 - 2
5x = -10
x = -10/5
x = -2

solution is (-2,4)
==================
36. 4x + 6y = 2000
       x + y = 410
==================
37. 4x + 6y = 2000
       x + y = 410......multiply by -6
----------------------
4x + 6y = 2000
-6x - 6y = - 2460 (result of multiplying by -6)
--------------------add
-2x = - 460
x = -460/-2
x = 230 <=== tickets in advance
 

3 0
3 years ago
Solve the following system
scZoUnD [109]

Answer:

{x = -4 , y = 2 ,  z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer:  {x = -4 , y = 2 ,  z = 1

4 0
3 years ago
What does 1/10 of 3000 equal?
arsen [322]
Answer:-
1/10×3000
=300
4 0
3 years ago
Read 2 more answers
Someone please help me
Lilit [14]
<h3>Answer:</h3>
  1. C. (9x -1)(x +4) = 9x² +35x -4
  2. B. 480
  3. A. P(t) = 4(1.019)^t

Step-by-step explanation:

1. See the attachment for the filled-in diagram. Adding the contents of the figure gives the sum at the bottom, matching selection C.

2. If we let "d" represent the length of the second volyage, then the total length of the two voyages is ...

... (d+43) + d = 1003

... 2d = 960 . . . . . . . subtract 43

... d = 480 . . . . . . . . divide by 2

The second voyage lasted 480 days.

3. 1.9% - 1.9/100 = 0.019. Adding this fraction to the original means the original is multiplied by 1 +0.019 = 1.019. Doing this multiplication each year for t years means the multiplier is (1.019)^t.

Since the starting value (in 1975) is 4 (billion), the population t years after that is ...

... P(t) = 4(1.019)^t

7 0
3 years ago
Consider two people being randomly selected. (For simplicity, ignore leap years.)
inna [77]

Answer:

(a) = \frac{144}{133225} \\\\(b) = \frac{1}{365}

Step-by-step explanation:

Part (a) the probability that two people have a birthday on the 9th of any month.

Neglecting leap year, there are 365 days in a year.

There are 12 possible 9th in months that make a year calendar.

If two people have birthday on 9th; P(1st person) and P(2nd person).

=\frac{12}{365} X\frac{12}{365}  = \frac{144}{133225}

Part (b) the probability that two people have a birthday on the same day of the same month

P(2 people selected have birthday on the same day of same month) + P(2 people selected not having birthday on  same day of same month) = 1

P(2 people selected not having birthday on  same day of same month):

= \frac{365}{365} X \frac{364}{365} =\frac{364}{365}

P(2 people selected have birthday on the same day of same month) = 1-\frac{364}{365} \\\\= \frac{1}{365}

7 0
3 years ago
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