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snow_tiger [21]
2 years ago
6

The parking garage at the airport has 72 empty parking spaces and 28 full parking spaces.

Mathematics
2 answers:
Mila [183]2 years ago
7 0

Answer:

72% because 72+28 is 100 and 72/100 is 72%

Step-by-step explanation:

viktelen [127]2 years ago
3 0

Answer:

72% of the parking spaces are empty.

Step-by-step explanation:

There are 72 open spaces out of 100 total. 72/100=.72, .72*100= 72%.

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100deciliters equals 10 liters

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3 years ago
What is Sally’s average speed? Plz answer ASAP
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5 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
Determine the value of variables a, b, and c that make each equation true.
dybincka [34]

Corrected Question

Determine the values of a, b and c that make each equation true.

(x^a)^6=\dfrac{1}{x^{30}} \\\\(x^{-7})^{-4}=x^b\\\\(x^{-2})^c=x^{22}

Answer:

a=-5, b=28 and c=-11

Step-by-step explanation:

To solve for a,b and c, we apply the following laws of indices

\dfrac{1}{x^y}=x^{-y} \\\\(x^m)^n=x^{m X n}\\\\$If x^m=x^n,$ then m=n

Therefore

(x^a)^6=\dfrac{1}{x^{30}}\\x^{a*6}=x^{-30}\\6a=-30\\a=-5

To solve for b

(x^{-7})^{-4}=x^b\\x^{-7*-4}=x^b\\x^{28}=x^b\\b=28

To solve for c

(x^{-2})^c=x^{22}\\x^{-2*c}=x^{22}\\-2c=22\\c=-11

4 0
3 years ago
Complete the solution of the equation. Find the<br> value of y when x equals 10.<br> -7x - 5y = -60
belka [17]

Answer:

y = 2

Step-by-step explanation:

-7x - 5y = -60

plug in 10 where x is

-7(10) - 5y = -60

multiply -7 and 10

-70 - 5y = -60

add 70 to both sides

-5y = 10

divide both sides by -5

y = -2

Hope this helps, if not let me know.

6 0
3 years ago
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