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Fittoniya [83]
3 years ago
7

How do I do the bus stop method?

Mathematics
1 answer:
hichkok12 [17]3 years ago
6 0
The bus stop method is just like division.
Ex
137÷5= 
1.) 5 goes into one zero times so there is a zero on top.
2.) 5 goes into 13 two times, so after the zero there is a 2.
3.) 5 goes into 37 seven times so there is a 7 after the 2.
4.) There is a 2 left and five can't go into school therefore that is a remainder.

    027R2
5/137
  <u>- 10</u>
     37
   <u>-35</u>
       2
Hoped I helped:D

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Pre-Calculus - Systems of Equations with 3 Variables please show work/steps
inessss [21]

Answer:

x = 10 , y = -7 , z = 1

Step-by-step explanation:

Solve the following system:

{x - 3 z = 7 | (equation 1)

2 x + y - 2 z = 11 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Swap equation 1 with equation 2:

{2 x + y - 2 z = 11 | (equation 1)

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

-x - 2 y + 9 z = 13 | (equation 3)

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

{2 x + y - 2 z = 11 | (equation 1)

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

-x - 2 y + 9 z = 13 | (equation 3)

Multiply equation 2 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

-x - 2 y + 9 z = 13 | (equation 3)

Add 1/2 × (equation 1) to equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - (3 y)/2 + 8 z = 37/2 | (equation 3)

Multiply equation 3 by 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - y - 4 z = 3 | (equation 2)

0 x - 3 y + 16 z = 37 | (equation 3)

Swap equation 2 with equation 3:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x - y - 4 z = 3 | (equation 3)

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

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

0 x+0 y - (28 z)/3 = (-28)/3 | (equation 3)

Multiply equation 3 by -3/28:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y + 16 z = 37 | (equation 2)

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

Subtract 16 × (equation 3) from equation 2:

{2 x + y - 2 z = 11 | (equation 1)

0 x - 3 y+0 z = 21 | (equation 2)

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

Divide equation 2 by -3:

{2 x + y - 2 z = 11 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

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

Subtract equation 2 from equation 1:

{2 x + 0 y - 2 z = 18 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

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

Add 2 × (equation 3) to equation 1:

{2 x+0 y+0 z = 20 | (equation 1)

0 x+y+0 z = -7 | (equation 2)

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

Divide equation 1 by 2:

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

0 x+y+0 z = -7 | (equation 2)

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

Collect results:

Answer: {x = 10 , y = -7 , z = 1

3 0
3 years ago
160 bananas;20%decrease
Jobisdone [24]

Answer:

128 banana's... I would work it all out for u but the teacher just came in my room and had a go

7 0
2 years ago
Read 2 more answers
Find the sum of 3b/b 2 and 12/b 2 and express it in simplest form
alexandr1967 [171]
\frac{3b}{b^{2}} + \frac{12}{b^{2}} = \frac{3b + 12}{b^{2}} = \frac{3(b) + 3(4)}{b^{2}} = \frac{3(b + 4)}{b^{2}}
8 0
3 years ago
Given: <br> PQ<br> ⊥<br> QR<br> , PR=20,<br> SR=11, QS=5<br> Find: The value of PS.
dlinn [17]

Answer:

The value of the side PS is 26 approx.

Step-by-step explanation:

In this question we have two right triangles. Triangle PQR and Triangle PQS.

Where S is some point on the line segment QR.

Given:

PR = 20

SR = 11

QS = 5

We know that QR = QS + SR

QR = 11 + 5

QR = 16

Now triangle PQR has one unknown side PQ which in its base.

Finding PQ:

Using Pythagoras theorem for the right angled triangle PQR.

PR² = PQ² + QR²

PQ = √(PR² - QR²)

PQ = √(20²+16²)

PQ = √656

PQ = 4√41

Now for right angled triangle PQS, PS is unknown which is actually the hypotenuse of the right angled triangle.

Finding PS:

Using Pythagoras theorem, we have:

PS² = PQ² + QS²

PS² = 656 + 25

PS² = 681

PS = 26.09

PS = 26

7 0
3 years ago
50 POINTS!!!
san4es73 [151]

Step-by-step explanation:

<u>Step 1:  Determine an ordered pair</u>

A solution of an equation just means that the point lies on the line.  We can find any y-value when we plug in a specific x-value.  For example, if we want to know what ordered pair lies at x=1, we just plug in y = -1/2(1) and solve for y which gives us -1/2.  This gives us an ordered pair of (1, -1/2).  We can continue to do this for any x value.

We can also reverse the order and plug in the y-value and get the x-value in order to accomplish the same goal but it's a bit harder.

Hope this helps!

6 0
2 years ago
Read 2 more answers
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