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fredd [130]
4 years ago
12

In triangle ABC, the measure of angle B is 90 degrees, BC=16, and AC=20. triangle DEF is similar to triangle ABC, where vertices

D, E, and F corresponds to vertices A, B, and C, respectively, and each side of triangle DEF is 1/3 the length of the corresponding side of triangle ABC. what is the value of sinF?
(please show your work)
Mathematics
1 answer:
kkurt [141]4 years ago
6 0
 <span>The third side of triangle ABC is AB. Using the Pythagorean Theorem, its length is 12. 
12² + 16² = 20² 

∠F is congruent to ∠C and so the sin(∠F) = sin(∠C) 

The sin(∠C) = opposite/hypotenuse 
= |AB| / |AC| 
= 12/20 
= 3/5 
= 0.6 

So the answer is 0.6.</span>
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Answer:

Step-by-step explanation:

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y = 3x - 6

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The y-intercept is "b" so in this case the y-intercept is -6

Step 3: Use  \frac{rise}{run} to graph this equation

We start at (0,3) the y-intercept and go up by 3 and left by 1 (\frac{rise}{run} =\frac{3}{1})

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Answer Part A and Part B for the triangle shown
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Answer:

for Part A

Answer is C.

Step-by-step explanation:

\tan(a)  =  \frac{opposite}{adjacent }  \\

Here the angle, a is 60°, Opposite is 20 and adjacent is x.

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\tan(60 )  =  \frac{20}{x}

Making x the subject gives us

x =  \frac{20}{ \tan(60) }

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3 years ago
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:

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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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