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sergij07 [2.7K]
3 years ago
6

Please help asap 25 pts

Mathematics
2 answers:
o-na [289]3 years ago
5 0

HI There!

Solve: 3(x + 4) <= 36

--------------------------------------------

Solution:

3(x + 4) ≤ 36

3x + 12 ≤ 36

3x + 12 − 12 ≤ 36 − 12

3x ≤ 24

3x/3 ≤ 24/3

x ≤ 8

--------------------------------------------

Answer: B). x ≤ 8

--------------------------------------------

Hope This Helps :)

erma4kov [3.2K]3 years ago
4 0

3(x+4)\leq36\\\\3x+12\leq36\\\\3x\leq24\\\\x\leq8

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Answer:

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3 years ago
Solve the equation.<br> n+ 6.3 = 16.3<br> O 10<br> 0-10<br> O 22.6<br> 0 -22.6
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Answer:

10

Step-by-step explanation:

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3 years ago
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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
4 years ago
Simplify to create an equivalent expression 8(10-6q)+3(-7q-2)
Temka [501]

Answer:

74-69q

Step-by-step explanation:

8(10-6q)+3(-7q-2)\\\\80-48q-21q-6\\\\80-69q-6\\\\80-6-69q\\\\\boxed{74-69q}

This can also be -69q + 74.

Hope this helps.

4 0
4 years ago
Consider circle T with radius 24 in. And 0=startfraction 5 pi over 6 endfraction radians. Circle T is shown. Line segments S T a
satela [25.4K]

Answer:

20\pi $ inches or \approx 62.83$ inches

Step-by-step explanation:

\text{Central Angle of Arc SV}=\dfrac{5\pi}{6}$ rad$\\\text{Radius of Circle T}=24 in.\\\\\text{Length of an arc} = \dfrac{\theta}{2\pi} \times 2\pi r

Therefore:

\text{Length of minor arc SV} = \dfrac{\frac{5\pi}{6}}{2\pi} \times 2\pi \times 24\\=20\pi $ inches\\\approx 62.83$ inches

The length of minor arc SV is 62.83 inches.

7 0
3 years ago
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