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yanalaym [24]
3 years ago
14

WILL MARK BRAINLIEST

Mathematics
1 answer:
Leto [7]3 years ago
7 0

Please mark me brainliest! :D

Part 1) To find the measure of ∠A in ∆ABC, use

we know that

In the triangle ABC

Applying the law of sines

in this problem we have

therefore

the answer Part 1) is

Law of Sines

Part 2) To find the length of side HI in ∆HIG, use

we know that

In the triangle HIG

Applying the law of cosines

In this problem we have

g=HI

G=angle Beta

substitute

therefore

the answer Part 2) is

Law of Cosines

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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
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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).

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Answer:

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Step-by-step explanation:

6(x+y)-5

Distribute using the distributive property

6x+6y-5

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