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MrRissso [65]
3 years ago
8

Help!!! Need help. I can’t fail this!! 2222

Mathematics
2 answers:
mixas84 [53]3 years ago
8 0

Answer:

c

Step-by-step explanation: because the one on the bottom is smaller than the other

Tasya [4]3 years ago
7 0

Simple. It’s the first one.

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Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for th
Masteriza [31]

Answer:

Let t_i be the time for the ith task.

We know these times have a certain structure:

  • Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5  \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6

  • The second task takes 1 second t_2=1
  • The fourth task takes 10 seconds t_4=10

So, we have the following system of equations:

t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0  \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

  • Subtract row 2 from row 1 \left(R_1=R_1-R_2\right).
  • Subtract row 2 from row 3 \left(R_3=R_3-R_2\right).
  • Add row 3 to row 2 \left(R_2=R_2+R_3\right).
  • Multiply row 3 by −1 \left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right).
  • Add row 4 multiplied by \frac{3}{2} to row 1 \left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right).
  • Subtract row 4 from row 3 \left(R_3=R_3-R_4\right).

Here is the reduced echelon form for the augmented matrix.

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]

c) The additional rows are

\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right

and the augmented matrix is

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]

d) To solve the system you must use these row operations.

  • Subtract row 1 from row 6 \left(R_6=R_6-R_1\right).
  • Subtract row 2 from row 6 \left(R_6=R_6-R_2\right).
  • Subtract row 3 from row 6 \left(R_6=R_6-R_3\right).
  • Swap rows 5 and 6.
  • Add row 5 to row 3 \left(R_3=R_3+R_5\right).
  • Multiply row 5 by 2 \left(R_5=\left(2\right)R_5\right).
  • Subtract row 6 multiplied by 1/2 from row 1 \left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right).
  • Add row 6 multiplied by 1/2 to row 3 \left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right).

\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]

The solutions are: (t_1,...,t_6)=(5,1,44,10,90,20).

5 0
3 years ago
Given V = 2πr2h, solve for h
salantis [7]
H= 2v/ pier^2
height equals two times the volume divided by pie( rsquared)
7 0
4 years ago
Read 2 more answers
How do I solve this Geometry problem?
tia_tia [17]

Answer:

you let other people do it also all in totle is 50,8 oe 3

Step-by-step explanation:

5 0
3 years ago
Fine 15/4 multiple by the reciprocal of 5/4
son4ous [18]

Answer:

3

Step-by-step explanation:

15/4×4/5

60/20

3

6 0
4 years ago
Read 2 more answers
What total length of ribbon is needed to Line the two long sides of the hat?
frozen [14]

Answer:

Total length of ribbon to line the long sides will be 22 in

Step-by-step explanation:

In the given figure of kite if we draw a line "h" vertically will form two right angle triangles.

Then by applying Pythagoras theorem in both the triangles.

h²= (3x + 5)² + (2x + 1)² ------(1)

h²= (5x + 1)² + 5² ------(2)

By equating both the equations

(3x + 5)² + (2x + 1)² = (5x + 1)² + 25

9x² + 25 + 30x + 4x² + 4x + 1 = 25x² + 10x + 1 + 25

13x² + 34x + 26 = 25x² + 10x + 26

13x² - 25x² + 34x - 10x + 26 - 26 = 0

- 12x² + 24x = 0

12x² - 24x = 0

x² - 2x = 0

x(x - 2) = 0

x = 2

Now we will put x = 2 in the measure of sides of the kite.

Side 1 = (3x + 5)

          = 3×2 + 5

          = 11

Side 2 = (5x + 1)

          = 5×2 + 1

          = 11

Side 3 = (2x + 1)

          = 2×2 + 1

          = 5

Therefore, Total length of ribbon to line the long sides will be = 11 + 11

= 22 in.

6 0
4 years ago
Read 2 more answers
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