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seropon [69]
2 years ago
7

a division problem the ____ is the original number, which is being divided by the divisor, to get the quotient.

Mathematics
1 answer:
ohaa [14]2 years ago
6 0

Answer:

dividend

Step-by-step explanation:

The DIVIDEND is the original number, which is being divided by the divisor:

                    QUOTIENT

                 ------------------------                

DIVISOR   )    DIVIDEND

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Can I get help on this please? You will be marked Brainliest and 5 ⭐️ <br> Thank you!!!
miskamm [114]

Answer:

y = 50x

Step-by-step explanation:

the equation for this is y = 50x because for every 50 units we go up by on the graph we go 1 unit to the right so:

slope-intercept form: y = mx + b   where m is the slope and b is the y-intercept. Our y-intercept is 0 because that is where our line intersects the y-axis and because it is 0 we do not need to put it in the equation.

y = 50x

7 0
3 years ago
Read 2 more answers
Solve y=-x+2 and y=-2x-1 as a graph, substitution, or elimination?
kotegsom [21]

Answer:

(- 3, 5 )

Step-by-step explanation:

Given the 2 equations

y = - x + 2 → (1)

y = - 2x - 1 → (2)

Using substitution method.

Substitute y = - x + 2 into (2)

- x + 2 = - 2x - 1 ( add 2x to both sides )

x + 2 = - 1 ( subtract 2 from both sides )

x = - 3

Substitute x = - 3 into (1)

y = - (- 3) + 2 = 3 + 2 = 5

Solution is (- 3, 5 )

8 0
3 years ago
I'm having trouble
Liono4ka [1.6K]
It would be B because you would take amelias age and take two away since todd is five years younger to get the a-5.
hope this helps :)
8 0
3 years ago
I can't figure out how to do <br> -5+x/3=-11
-BARSIC- [3]
Add 5 to both sides
5-5+x/3=5-11
0+x/3=-6
x/3=-6
times both sides by 3
3x/3=-18
x=-18
3 0
3 years ago
This is Matrix for pre calc
gavmur [86]

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

  • Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 1 by 1/3:

\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

6 0
3 years ago
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