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Temka [501]
3 years ago
10

Can anyone help me. Will give you all my points and brainliest:)

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
8 0

See the attachments for the answer :)

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Look at the picture! Express the following set in set-builder notation ?
Alex

Answer:

A.   E={x|x\in N and x is a multiple of 8}

Step-by-step explanation:

Let x represent the set of natural numbers. Then we can write;

x\in N.

The set of all natural numbers that are multiples of 8 is then written as;

x\in N and x is a multiple of 8.

E is the set of natural numbers that are multiples of 8 can then be written in set builder notation as;

E={x|x\in N and x is a multiple of 8}

The correct choice is A

7 0
3 years ago
the equation y=0.5x represents the distance henry hikes, in miles, over time, in hours. the graph represents the rate that clark
zheka24 [161]

I think there's supposed to be a picture here. Anyway, you can find out by finding the slope of the line on the graph and comparing it to .5 which is Henry's slope. If the graphed slope is larger than .5 then Clark hikes faster, if it is less than .5, Henry hikes faster.

Hope this helps :)

6 0
3 years ago
3(x -y) + 4 if X = 3 and y = 2
Alchen [17]

Answer:

7

Step-by-step explanation:

3(3-2)+4

3(1)+4

3+4

7

7 0
2 years ago
How to simplify 6 to the power of -1 over -8 to the power of 0
neonofarm [45]

Answer:

\frac{1}{6}

-8 to the power of 0 = 1

(Anything to the power of 0 = 1)

and 6 to the power of -1 = 1/6

( Any number to the power of a negative number will become a fraction)

Example.

5^{ - 3}  =  \frac{1}{5^{3} }

So is will be 1/6 over 1

Which gives you your final answer.

4 0
2 years ago
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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