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creativ13 [48]
3 years ago
11

What is 4(2×-7)+3(3×-1)=3(2-×)​

Mathematics
2 answers:
Mama L [17]3 years ago
8 0

Answer:

x=37/20

Step-by-step explanation:

4(2x-7)+3(3x-1)=3(2-x)

8x-28+9x-3=6-3x

17x+3x=6+31

20x=37

fenix001 [56]3 years ago
8 0

Answer:

x = \frac{71}{3}

Step-by-step explanation:

1. Simplify/distribute each side.

4(2)(-7)+3(3)(-1)=3(2-x)

4(2)(-7)+3(3)(-1)=(3)(2)+(3)(-x)

-56+-9=6+-3x

2. Flip the equation around.

-3x+6=-65

3. Subtract 6 to balance the equation.

-3x+6-6=-65-6

-3x=-71

4. Divide each side by 3. We know they are in the same fact family.

\frac{-3x}{-3} = \frac{-71}{3}

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10. What is the value of x?<br> 105°<br> 148°<br> x
Ostrovityanka [42]

Answer:

Step-by-step explanation:

The supplement of 105 = 180 - 105 = 75

The supplement of 148 = 180 -148 = 32

x is the sum of these two angles

x = 32 + 75

x = 107

Exterior angles are the sum of the 2 angles not connect to the exterior angle you are trying to find.

4 0
3 years ago
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C or D

Step-by-step explanation:

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How to find the measure of an angle?
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You can use a protractor

Step-by-step explanation:

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3 years ago
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Find the reduced row echelon form of the following matrices and then give the solution to the system that is represented by the
GaryK [48]

Answer:

a)

Reduced Row Echelon:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

Solution to the system:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

Reduced Row Echelon:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

Solution to the system:  

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

Step-by-step explanation:

To find the reduced row echelon form of the matrices, let's use the Gaussian-Jordan elimination process, which consists of taking the matrix and performing a series of row operations. For notation, R_i will be the transformed column, and r_i the unchanged one.

a) \left[\begin{array}{cccc}0&4&7&0\\2&1&0&0\\0&3&1&-4\end{array}\right]

Step by step operations:

1. Reorder the rows, interchange Row 1 with Row 2, then apply the next operations on the new rows:

R_1=\frac{1}{2}r_1\\R_2=\frac{1}{4}r_2

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right]

2. Set the first row to 1

R_3=-3r_2+r_3

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

3. Write the system of equations:

x_1+\frac{1}{2}x_2=0\\x_2+\frac{7}{4}x_3=0\\x_3=-4

Now you have the  reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right]

1. R_2=-2r_1+r_2\\R_3=-r_1+r_3

Resulting matrix:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

2. Write the system of equations:

4x_1+3x_2=7\\2x_3=-17

Now you have the reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

7 0
3 years ago
Will give brainliest answer
Ann [662]

Answer:

not equivalent

equivalent

not equivalent

Step-by-step explanation:

25 is by itself already 5²

therefore

{25}^{m}  =  {5}^{2m}

when we divide one time by 5, we simply take away 1 from the power making it

{5}^{2m - 1}

the other options are wrong

{25}^{m - 1}

would be right, if we have

{25}^{m}  \div 25

but we don't.

and

{25}^{2m - 1}

would even square

{25}^{m}

and then divide by 25. no, the original excision is nothing like that.

7 0
3 years ago
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