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Elanso [62]
3 years ago
7

0.02 is 10 times 0.2

Mathematics
1 answer:
atroni [7]3 years ago
6 0
Yes that answer is correct
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If JL=18. Then which statement is true?
kondor19780726 [428]

Answer:

JK = 18 also

Step-by-step explanation:

3 0
3 years ago
1/4+2/3-1/2= how do I solve this
Ann [662]

Answer:

5/12

Step-by-step explanation:

First you add 1/4+2/3.

You need to change the denominator so that they are equal to each other

If you multiply the 1/4 by 3/3 you will get 3/12 which is equal to 1/4

Then you need to do the same to 2/3. This time you need to multiply it by 4/4 to get the same denominator which will be 2/3*4/4=8/12

Then you add 3/12+8/12=11/12. You don't need to add the denominator.

After this you will need to subtract 11/12 and 1/2. This time you need to only change the denominator for 1/2.

You multiply the denominator and numerator by 6/6 to 1/2 and you will get 6/12 then you are going to subtract 11/12-6/12 and you will get 5/12

6 0
3 years ago
Find m S<br> 12x +4<br> G<br> 40°<br> F<br> E<br> D<br> 4x
NISA [10]

Answer: C

Step-by-step explanation:

8 0
2 years ago
HELP DO NOT UNDERSTAND THIS QUESTION AT ALL THE BEST ONE WILL BE MARKED AS BRAINLIEST
Murrr4er [49]

Step-by-step answer:

A line with slope (or gradient) m, passing through a point (x0,y0) can be represented by the equation:

(y-y0) = m(x-x0)  ......................(1)

Here,

(x0,y0) = (1,7), and

slope/gradient = -2

Substitute values into (1) above gives

(y-7)=2(x-1)

Expand and simplify

y-7 = 2x-2

y =2x -2 + 7

y=2x+5

is the final answer.

5 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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