The minute hand of a clock is 9.5cm long. How far will the hand travel in one day, to the nearest meter

Answer:

To solve the system of linear equations using Gaussian elimination, we'll write the augmented matrix and perform row operations to transform it into row echelon form.

The given system of equations:

-2x_1 + 3x_2 + x_3 = 2

-3x_1 + 4x_2 + 2x_3 = 2

x_1 - 5x_2 + 4x_3 = -9

-2x_1 + 4x_2 - 4x_3 = 8

Writing the augmented matrix:

[ -2 3 1 | 2 ]

[ -3 4 2 | 2 ]

[ 1 -5 4 | -9 ]

[ -2 4 -4 | 8 ]

1. Row 1 Ã (-3) + Row 2 â Row 2:

[ -2 3 1 | 2 ]

[ 9 -13 -5 | -6 ]

[ 1 -5 4 | -9 ]

[ -2 4 -4 | 8 ]

2. Row 1 Ã (1/2) + Row 3 â Row 3:

[ -2 3 1 | 2 ]

[ 9 -13 -5 | -6 ]

[ -1 (3/2) 2 | -5/2]

[ -2 4 -4 | 8 ]

3. Row 1 Ã (-1) + Row 4 â Row 4:

[ -2 3 1 | 2 ]

[ 9 -13 -5 | -6 ]

[ -1 (3/2) 2 | -5/2]

[ 0 7 -3 | 6 ]

4. Row 2 Ã (-9/7) + Row 3 â Row 3:

[ -2 3 1 | 2 ]

[ 9 -13 -5 | -6 ]

[ -1 0 (59/7) | -(23/7)]

[ 0 7 -3 | 6 ]

5. Row 2 Ã (2/3) + Row 4 â Row 4:

[ -2 3 1 | 2 ]

[ 9 -13 -5 | -6 ]

[ -1 0 (59/7) | -(23/7)]

[ 0 0 (-11/7) | 0 ]

6. Row 3 Ã (-14/11) + Row 4 â Row 4:

[ -2 3 1 | 2 ]

[ 9 -13 -5 | -6 ]

[ -1 0 (59/7) | -(23/7)]

[ 0 0 1 | 0 ]

7. Row 3 Ã (1/2) + Row 1 â Row 1:

[ -1 3/2 3/2 | (4/7) ]

[ 9 -13 -5 | -6 ]

[ -1 0 1 | 0 ]

[ 0 0 1 | 0 ]

8. Row 3 Ã (9/7) + Row 2 â Row 2:

[ -1 3/2 3/2 | (4/7) ]

[ 0 -26/7 -14/7 | -42/7 ]

[ -1 0 1 | 0 ]

[ 0 0 1 | 0 ]

9. Row 3 Ã (1/2) + Row 4 â Row 4:

[ -1 3/2 3/2 | (4/7) ]

[ 0 -26/7 -14/7 | -42/7 ]

[ -1 0 1 | 0 ]

[ 0 0 1 |

Answer:

he ded

Step-by-step explanation:

\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \tohe no alive  because ⇆ω⇆π⊂∴∨α∈\neq  \lim_{n \to \infty} a_n \pi \left \{ {{y=2} \atop {x=2}} \right. \leq \neq \beta \beta \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \geq \geq \leq \leq \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right.  \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \beta x_{12 \lim_{n \to

Answer:

(x - 10)(x - 5)

Step-by-step explanation:

Well to factor X^2-15x+50

we need to find 2 numbers that multiply to get 50 and add to get -15.

-5 * -10 = 50

-5 + -10

x*x = x^2

Factored (x - 10)(x - 5)

Hope this helps :)

Answer:

(x - 10)(x - 5)

Step-by-step explanation:

Step 1- Find 2 numbers that multiply to be 50.

10 × 5

-10 × -5

25 × 2

-25 × -2

All these multiply to get 50.

But the numbers must also add up to be -15.

Step 2- Find the 2 numbers that also add up to be -15.

10 + 5 = 15

-10 + -5 = -15

25 + 2 = 27

-25 + -2 = -27

The correct equation would be - 10 + -5

The 2 number you would use when you factor would be -10 and -5

Step 3- Write the equation

You can write the equation as (x - 10)(x - 5)

Or you can also write it as (x - 5)(x - 10)

Answer:

Expected number of students  would have a driver's license = 6

Step-by-step explanation:

Explanation:-

Given data In a class of 25 students, 15 of them have a driver's license

The sample proportion

Let 'X' be the binomial distribution

Given sample size 'n' = 10

mean of the binomial distribution or expected number of students  would have a driver's license

                                 μ = n p

                                    = 10 × 0.6

                                   = 6

Conclusion:-

Expected number of students  would have a driver's license = 6

Answer:

The number of dolls sold 10,400

Step-by-step explanation:

Given:  The number N of dolls sold varies directly with their advertising budget A and inversely with the price P of each doll.

N = k(A/P), where k is the constant.

Now we have to find k.

Given: N = 5200, A = 26,000 and P = 30

5200 = k (26000/30)

5200 = k(866.67)

k = 5.99, when we round off we get k = 6

Now let's find the number of dolls sold when the ad amount increase to $52,000

Now plug k = 6, A = 52000 and p = 30

N = 6(52000/30)

N = (52000/5)

N = 10.400

Therefore, the number of dolls sold 10,400

Hope this will helpful.

Thank you.