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Naily [24]
3 years ago
9

Maria is using the random number table shown to select students for her survey. She assigned each of the 300 students in the sch

ool a three-digit number starting with 001. Beginning on Line 13, what is the third student number chosen for the sample?
A. 020
B. 171
C. 529
D. 639

Mathematics
1 answer:
pishuonlain [190]3 years ago
8 0
I believe the answer is C
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Which is the additive inverse of the complex number -8+3i?
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The additive inverse of a complex z is a complex number -z, so that

\mathsf{z+(-z)=-z+z=0+0i=0}


Finding -z:

\mathsf{z=-8+3i}\\\\\\ \mathsf{z+(-z)=0+0i}\\\\\ \mathsf{-8+3i+(-z)=0+0i}\\\\ \mathsf{-z=0+0i+8-3i}\\\\ \mathsf{-z=(0+8)+(0i-3i)}\\\\ \boxed{\begin{array}{c}\mathsf{-z=8-3i} \end{array}}\qquad\checkmark


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Tags: <em>complex number additive inverse opposite algebra</em>

3 0
3 years ago
How to classify 8x4 + 7x3 + 5x2 + 8 by number of terms.
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Answer:

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On a single set of axes, sketch a picture of the graphs of the following four equations: y = −x+ √ 2, y = −x− √ 2, y = x+ √ 2, a
Artist 52 [7]

Answer:

( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 ),  ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )  

y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 ,  y + 1 = ( x + 2 ) - √ 2

             y + 1 = ( x + 2 ) + √ 2  ,   ( x + 2 )^2 + ( y + 1)^2 = 1

Step-by-step explanation:

Given:

- Four functions to construct a diamond:

                y = −x+ √ 2,  y = −x− √ 2,  y = x+ √ 2, and y = x − √ 2.

Find:

a)Show that the unit circle sits inside this diamond tangentially; i.e. show that the unit circle intersects each of the four lines exactly once.

b)Find the intersection points between the unit circle and each of the four lines.

(c) Construct a diamond shaped region in which the circle of radius 1 centered at (−2, − 1) sits tangentially. Use the techniques of this section to help.

Solution:

- For first part see the attachment.

- The equation of the unit circle is given as follows:

                                      x^2 + y^2 = 1

- To determine points of intersection we have to solve each given function of y with unit circle equation for set of points of intersection:

                                For:  y = −x+ √ 2 , x - √ 2

                                And: x^2 + y^2 = 1

                                x^2 + (+/- * (x - √ 2))^2 = 1

                                x^2 + (x - √ 2)^2 = 1

                                2x^2 -2√ 2*x + 2 = 1

                                2x^2 -2√ 2*x + 1 = 0

                                 2[ x^2 - √ 2] + 1 = 0

Complete sqr:         (1 - 1/√ 2)^2 = 0

                                 x = 1/√ 2 , x = 1/√ 2                                          

                                 y = -1/√ 2 + √ 2 = 1/√ 2

                                 y = 1/√ 2 - √ 2 = - 1/√ 2

Points are:                ( 1/√ 2 , 1/√ 2 ) , ( 1/√ 2 , - 1/√ 2 )

- Using vertical symmetry of unit circle we can also evaluate other intersection points by intuition:

                                x = - 1/√ 2

                                 y = 1/√ 2 , -1/√ 2

Points are:              ( -1/√ 2 , 1/√ 2 ) , ( -1/√ 2 , - 1/√ 2 )  

- To determine the function for the rhombus region that would be tangential to unit circle with center at ( - 2 , - 1 ):

- To shift our unit circle from origin to ( - 2 , - 1 ) i.e two units left and 1 unit down.

- For shifts we use the following substitutions:

                           x = x + 2  ....... 2 units of left shift

                           y = y + 1 .......... 1 unit of down shift

- Now substitute the above shifting expression in all for functions we have:

                          y = −x+ √ 2 ----->  y + 1 = - ( x + 2 ) + √ 2

                          y = −x− √ 2 ----->  y + 1 = - ( x + 2 ) - √ 2

                          y = x- √ 2 ------->  y + 1 = ( x + 2 ) - √ 2

                          y = x+ √ 2 ------> y + 1 = ( x + 2 ) + √ 2

                          x^2 + y^2 = 1 ----->  ( x + 2 )^2 + ( y + 1)^2 = 1

- The following diamond shape graph would have the 4 functions as:

             y + 1 = - ( x + 2 ) + √ 2 , y + 1 = - ( x + 2 ) - √ 2 ,  y + 1 = ( x + 2 ) - √ 2

             y + 1 = ( x + 2 ) + √ 2  ,   ( x + 2 )^2 + ( y + 1)^2 = 1

- See attachment for the new sketch.            

7 0
3 years ago
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