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soldi70 [24.7K]
3 years ago
14

Please someone actually help with this I don't understand how.​

Mathematics
1 answer:
tigry1 [53]3 years ago
8 0

Answer:

Scatter plot

Step-by-step explanation:

Answer is 5

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Graph each circle given below right the center and radius of each circle
expeople1 [14]

Answer:

Center: (-3,-2)

Radius: √6

The graph is attached.

Step-by-step explanation:

The equation of the circle has the form:

(x -h)^{2}+(y-k)^{2}=r^{2}

Where (h,k) is the  point of the center of the circle and r is the radius of the circle.

The equation given in the problem is

(x +3)^{2}+(y+2)^{2}=6

 Therefore:

h=-3

k=-2

The center is at (-3,-2)

And the radius is:

r^2=6\\r=\sqrt{6}

Then, you can graph it has you can see in the image attached.

5 0
3 years ago
Read 2 more answers
Which state ment is true regarding the graphed function
jek_recluse [69]

Answer:

F(-2)= g(-2)

Step-by-step explanation:

F(-2)= g(-2), both function have the same points of intersect.

5 0
3 years ago
What is the volume of this cylinder?
azamat
Divide 1017,36 by pi and by the height (the inverse of the formula for the volume):

1017,36/3,14 = 324
324/9 = 36
6 0
3 years ago
Read 2 more answers
WHO CAN solve it Please !
Mariulka [41]

Answer:

a) True

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em>            </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha<em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

<em>            cos²∝ = 1-sin²∝</em>

<u><em>step(ii):-</em></u>

<em>Now the  integration</em>

<em>         </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha<em></em>

<em>                                      = </em>\int\limits^\pi _0 {cos\alpha } \, dx<em></em>

<em>Now, Integrating </em>

<em>                                  </em>= ( sin\alpha )_{0} ^{\pi }<em></em>

<em>                                = sin π - sin 0</em>

<em>                               = 0-0</em>

<em>                              = 0</em>

<u><em>Final answer:-</em></u>

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

<em></em>

<em></em>

5 0
2 years ago
What is 1.667 as a fraction
Solnce55 [7]

Step-by-step explanation:

hoped this helped have a nice day

4 0
3 years ago
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