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Leto [7]
2 years ago
12

Find the domain and range

Mathematics
1 answer:
julia-pushkina [17]2 years ago
7 0

Answer:

Domain: (-∞, ∞)
Range: [-2, ∞)

Step-by-step explanation:

The domain of any parabola is (-∞, ∞)
The range of this parabola is [-2, ∞) because the vertex is at (-2, -2).

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Review your answers to Question 2. What happens to the coordinates of a shape when the shape reflects about the x-axis? What hap
mixas84 [53]

Answer:

When the shape reflects about the x-axis, the x-coordinate remains unchanged but the y-coordinate becomes negative. When the shape reflects about the y-axis, the y-coordinate remains the same but the x-coordinate becomes negative.

Step-by-step explanation:

Sample answer from Edmentum

4 0
3 years ago
Why do the inequality signs stay the same in the last two steps of exercise 1
MrRissso [65]
The only reason an inequality would change would be because a number is divided or multiplied by a negative number. So if it stayed the same, then it would be because there was no division or multiplication by a negative number. Hope I helped :)
7 0
2 years ago
Read 2 more answers
Given: PRST square
yuradex [85]
There is no difficulty in this problem until you construct the figures. How can we do it is shown in the attached picture. After drawing PRST, from the point P, we can draw PMKD and later we can complete PMCT as a result. From this picture, we can see that the side of PMCT is also a. Then, the area of this square is A=a^{2}

6 0
2 years ago
How to solve this question
VMariaS [17]
<span>The variance method is as follows.

-Sum the squares of the values in data set, and then divide by the number of values in data set
- From that, subtract the square of the mean (add all values and divide by number of values in the data set)

Our variance is

<span>\displaystyle\sigma^2 = \frac{2^2 + 5^2 + m^2}{3} - \left(\frac{2 + 5 + m}{3}\right)^2

Since variance has to be 14, we set \sigma^2  = 14 and solve for m

14= \frac{4 + 25 + m^2}{3} - \left(\frac{7 + m}{3}\right)^2\ \Rightarrow \\ \\&#10;14 = \frac{29}{3} + \frac{1}{3}m^2 - \frac{1}{9}(7+m)^2 \\ \\&#10;14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{1}{9}(49 + 14m + m^2) \\ \\&#10;14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{49}{9}- \frac{14}{9}m- \frac{1}{9}m^2 \\ \\&#10;0 = \frac{-88}{9}  -\frac{14}{9}m + \frac{2}{9}m^2&#10;

quadratic formula


m = \displaystyle\frac{-b \pm \sqrt{b^2 -4ac}}{2a} \\&#10;m = \frac{-(-\frac{14}{9}) \pm \sqrt{\left(-\frac{14}{9}\right)^2 - 4(2/9)(-88/9)} }{2(2/9)} \\&#10;m = \frac{\frac{14}{9} \pm \sqrt{ \frac{196}{81} + \frac{704}{81} } }{\frac{4}{9} } \\&#10;m = \frac{\frac{14}{9} \pm \sqrt{ \frac{900}{81}  } }{\frac{4}{9} } \\&#10;m = \frac{\frac{14}{9} \pm \sqrt{ \frac{100}{9}  } }{\frac{4}{9} } \\&#10;m = \frac{\frac{14}{9} \pm \frac{10}{3}  }{\frac{4}{9} } \\&#10;m = 11, -4

-4 doesnt' work as it is not a positive integer

m = 11


</span></span>
5 0
2 years ago
Is 3/4 greater than 7/8?
yKpoI14uk [10]
3/4 = (2*3) / (2* 4) = 6/8

7/8.

Is 3/4 greater than 7/8? 
Is like asking is 6/8 greater than 7/8?  Because 3/4 is equivalent to 6/8.

The answer is No.
3 0
3 years ago
Read 2 more answers
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