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Aleks [24]
3 years ago
10

Need help for this homework please it’s about opposite and negative vector

Mathematics
1 answer:
IgorLugansk [536]3 years ago
4 0

Ok what is your question

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5/6-2/5 in simplest form
DiKsa [7]

Answer:

13/30

Step-by-step:

First, you find the common denominator, which is 30. For the first fraction, multiply the bottom by five. Since you do that, you have to multiple the top by 5 as well, giving you 25/30. For the second problem, you multiple the top and bottom by six giving you 12/30. 25 minus 12 is 13 so you get 13/30.

7 0
3 years ago
(1)<br> (2)<br> 790<br> + 172<br><br><br> What is the answer
VladimirAG [237]

Answer:

962

Step-by-step explanation:

8 0
3 years ago
Find the sum of (63−4)+(−23+9)=
frutty [35]

Answer:

45

Step-by-step explanation

Hope this helps :)

8 0
3 years ago
Read 2 more answers
In the morning, Jasmine biked to her grandma's house in 45 minutes. When she was biking back home in the evening it was foggy an
Mars2501 [29]
Let x = speed and y = distance

y = x * 45

y = (x - 4) * 70

45x = 70x - 280

-25x = -280

x = 11.2

Put it back into the first equation:

y = 11.2 x 45

y = 504miles

Hope this helps! Any questions let me know :)
7 0
3 years ago
Solve the following system of equation
krok68 [10]

Answer:

x = 6 , y = -4 , Z = -1

Step-by-step explanation:

Solve the following system:

{2 x + 3 y - Z = 1 | (equation 1)

3 x + y + 2 Z = 12 | (equation 2)

-3 + x + 2 y = -5 | (equation 3)

Express the system in standard form:

{2 x + 3 y - Z = 1 | (equation 1)

3 x + y + 2 Z = 12 | (equation 2)

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

Swap equation 1 with equation 2:

{3 x + y + 2 Z = 12 | (equation 1)

2 x + 3 y - Z = 1 | (equation 2)

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

Subtract 2/3 × (equation 1) from equation 2:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+(7 y)/3 - (7 Z)/3 = -7 | (equation 2)

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

Multiply equation 2 by 3/7:

{3 x + y + 2 Z = 12 | (equation 1)

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

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

Subtract 1/3 × (equation 1) from equation 3:

{3 x + y + 2 Z = 12 | (equation 1)

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

0 x+(5 y)/3 - (2 Z)/3 = -6 | (equation 3)

Multiply equation 3 by 3:

{3 x + y + 2 Z = 12 | (equation 1)

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

0 x+5 y - 2 Z = -18 | (equation 3)

Swap equation 2 with equation 3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+y - Z = -3 | (equation 3)

Subtract 1/5 × (equation 2) from equation 3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+0 y - (3 Z)/5 = 3/5 | (equation 3)

Multiply equation 3 by 5/3:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+0 y - Z = 1 | (equation 3)

Multiply equation 3 by -1:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y - 2 Z = -18 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Add 2 × (equation 3) to equation 2:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+5 y+0 Z = -20 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Divide equation 2 by 5:

{3 x + y + 2 Z = 12 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Subtract equation 2 from equation 1:

{3 x + 0 y+2 Z = 16 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{3 x+0 y+0 Z = 18 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Divide equation 1 by 3:

{x+0 y+0 Z = 6 | (equation 1)

0 x+y+0 Z = -4 | (equation 2)

0 x+0 y+Z = -1 | (equation 3)

Collect results:

Answer:  {x = 6 , y = -4 , Z = -1

6 0
3 years ago
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