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Sav [38]
3 years ago
13

Six times a number is 45 greater than the the product of the number and -3. Find the number.

Mathematics
1 answer:
blagie [28]3 years ago
6 0

Let us try to formulate the expressions one by one.

let the number be x.

Step 1:

six time the number is 6*x = 6x

Step 2:

product of the number and -3 is -3x

Step 3:

now let us make equation for it:

6x = -3x +45

adding 3x on the left side,

6x+3x = 45

9x = 45

dividing right side by 9

x=45/9 = 5

So the number is x=5



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How to do the inverse of a 3x3 matrix gaussian elimination.
nata0808 [166]

As an example, let's invert the matrix

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}

We construct the augmented matrix,

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

On this augmented matrix, we perform row operations in such a way as to transform the matrix on the left side into the identity matrix, and the matrix on the right will be the inverse that we want to find.

Now we can carry out Gaussian elimination.

• Eliminate the column 1 entry in row 2.

Combine 2 times row 1 with 3 times row 2 :

2 (-3, 2, 1, 1, 0, 0) + 3 (2, 1, 1, 0, 1, 0)

= (-6, 4, 2, 2, 0, 0) + (6, 3, 3, 0, 3, 0)

= (0, 7, 5, 2, 3, 0)

which changes the augmented matrix to

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 1 & 1 & 1 & 0 & 0 & 1 \end{array} \right]

• Eliminate the column 1 entry in row 3.

Using the new aug. matrix, combine row 1 and 3 times row 3 :

(-3, 2, 1, 1, 0, 0) + 3 (1, 1, 1, 0, 0, 1)

= (-3, 2, 1, 1, 0, 0) + (3, 3, 3, 0, 0, 3)

= (0, 5, 4, 1, 0, 3)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 5 & 4 & 1 & 0 & 3 \end{array} \right]

• Eliminate the column 2 entry in row 3.

Combine -5 times row 2 and 7 times row 3 :

-5 (0, 7, 5, 2, 3, 0) + 7 (0, 5, 4, 1, 0, 3)

= (0, -35, -25, -10, -15, 0) + (0, 35, 28, 7, 0, 21)

= (0, 0, 3, -3, -15, 21)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 3 & -3 & -15 & 21 \end{array} \right]

• Multiply row 3 by 1/3 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 5 & 2 & 3 & 0 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 3 entry in row 2.

Combine row 2 and -5 times row 3 :

(0, 7, 5, 2, 3, 0) - 5 (0, 0, 1, -1, -5, 7)

= (0, 7, 5, 2, 3, 0) + (0, 0, -5, 5, 25, -35)

= (0, 7, 0, 7, 28, -35)

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 7 & 0 & 7 & 28 & -35 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 2 by 1/7 :

\left[ \begin{array}{ccc|ccc} -3 & 2 & 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Eliminate the column 2 and 3 entries in row 1.

Combine row 1, -2 times row 2, and -1 times row 3 :

(-3, 2, 1, 1, 0, 0) - 2 (0, 1, 0, 1, 4, -5) - (0, 0, 1, -1, -5, 7)

= (-3, 2, 1, 1, 0, 0) + (0, -2, 0, -2, -8, 10) + (0, 0, -1, 1, 5, -7)

= (-3, 0, 0, 0, -3, 3)

\left[ \begin{array}{ccc|ccc} -3 & 0 & 0 & 0 & -3 & 3 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

• Multiply row 1 by -1/3 :

\left[ \begin{array}{ccc|ccc} 1 & 0 & 0 & 0 & 1 & -1 \\ 0 & 1 & 0 & 1 & 4 & -5 \\ 0 & 0 & 1 & -1 & -5 & 7 \end{array} \right]

So, the inverse of our matrix is

\begin{bmatrix}-3&2&1\\2&1&1\\1&1&1\end{bmatrix}^{-1} = \begin{bmatrix}0&1&-1\\1&4&-5\\-1&-5&7\end{bmatrix}

6 0
2 years ago
20 POINTS AND BRAINLIEST!
Advocard [28]

Answer:

<A =21

<B = 84

<c = 75

Step-by-step explanation:

<A =x

<B = 4x

<C =5x-30

The sum of A+B+C = 180

 x+ 4x + 5x-30 = 180

Combine like terms

10x -30 = 180

Add 30 to each side

10x-30+30 = 180+30

10x = 210

Divide by 10

10x/10 = 210/10

x= 21

Since x =21 we can find the value of the angles

Substituting x into the equations for angles A, B and C

<A =21

<B = 4x = 4*21 = 84

<C = 5x-30 = 5*21-30 = 105-30 = 75

7 0
3 years ago
Read 2 more answers
Marshall's mom buys 2 boxes of granola bars each week. each box contains 8 granola bars. if she continues buying 2 boxes each we
sukhopar [10]
She buys 2 boxes each week for a year (there is 52 weeks in a year) = 2 x 52 = 104 boxes per year
if each box contains 8 granola bars...and she buys 104 boxes, then she buys:
104 * 8 = 832 granola bars
6 0
3 years ago
Do anybody know this answer because I need help
Papessa [141]

Answer:

x=-7

Step-by-step explanation:

divide 12 on both sides and that equals -7

8 0
3 years ago
If a 24-kg mass stretches a spring 15 cm, what mass will stretch the spring 10 cm
natulia [17]

Answer:

<h2>16kg</h2>

Step-by-step explanation:

 This problem is borers on elasticity of materials.

according to Hooke's law,<em> "provided the elastic limit of an elastic material is not exceeded the the extension e is directly proportional to the applied force."</em>

F= ke

where F is the applied force in N

          k is the spring constant N/m

          e is the extension in meters

Given data

mass m= 24kg

extensnion=15cm in meters= \frac{15}{100}= 0.15m

we can solve for the spring constant k

we also know that the force F = mg

assuming g=9.81m/s^{2}

therefore

24*9.81=k*0.15\\235.44=k*0.15\\k=\frac{235.44}{0.15} \\k=1569.6N/m\\

We can use this value of k to solve for the mass that will cause an extension of  10cm= 0.1m

x*9.81=1569.6*0.1\\\\x= \frac{156.96}{9.81} \\\x= 16kg

5 0
3 years ago
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