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vlada-n [284]
3 years ago
13

Which values from the set -6, -4, -3, -1, 0, 2 satisfy this inequality?

Mathematics
1 answer:
Marat540 [252]3 years ago
8 0

Answer:

-6, -4, -3, and -1 only

Step-by-step explanation:

hope this helps!

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The equation of a circle is given below. x^{2}+(y-2.25)^{2} = \dfrac{196}{169}x 2 +(y−2.25) 2 = 169 196 ​ x, squared, plus, left
Lostsunrise [7]
  • Center =(h,k) = (0,2.25)
  • Radius = r = \frac{14}{13}

<u>Step-by-step explanation:</u>

Here we have following equation : x^{2}+(y-2.25)^{2} = \dfrac{196}{169}

We need to find the center & radius of this circle . Let's find out:

We know that , Equation of a circle is given by :

⇒ (x-h)^2+(y-k)^2=r^2   ........(1)

Here , (h,k) are the co-ordinates of center & r is the radius of circle.Collectively called as a circle with radius r and center at (h,k) . Let's frame given equation in question :

⇒  x^{2}+(y-2.25)^{2} = \frac{196}{169}

⇒  (x-0)^{2}+(y-2.25)^{2} = (\frac{14}{13})^2

On comparing this equation with equation (1) we get :

  • Center =(h,k) = (0,2.25)
  • Radius = r = \frac{14}{13}
6 0
3 years ago
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.please help me with this I will crown you brainlest
Marizza181 [45]
You can prove it by the SAS postulate which should be the first option
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3 years ago
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Help help hep math math
MrRissso [65]

300 {ft}^{2}

Step-by-step explanation:

d=20

r=10

π=3

a = \pi{r}^{2}

a = (3)  {(10)}^{2}

a = 300

7 0
2 years ago
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What is the approximate circumference of a circle with a radius of 6.5 meters? Use π ≈ 3.14. 13 meters 20.4 meters 40.8 meters 1
natka813 [3]
Circumference = 2πR = 2×3.14×6.5 = 40.82m
5 0
3 years ago
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PLEASE HELP!!!!!!!!!!
Oksanka [162]

Answer:

B

Step-by-step explanation:

Data set D does not contain the value 128, which is the median value.

Data set C does not contain the outlier value 91.

Data set A contains value 168, which does not show up on the plot.

The only remaining choice is B.

_____

In order, the data values of set B are ...

... 91, 114, 120, 126, 128, 128 134, 136, 139, 142, 152

The median value of these 11 is the 6th one: 128. The median values of the remaining two sets of 5 are 120 and 139, making these values the quartiles at the ends of the box. The value 91 is more than 1.5 times the IQR (19) below the 1st quartile, so is considered an outlier. (The cutoff is 120-1.5·19=91.5.)

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