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Vikki [24]
3 years ago
8

What is the smallest number of degrees needed to rotate a regular pentagon around it's center onto itself ?

Mathematics
1 answer:
Leviafan [203]3 years ago
6 0
The smallest degree that you can rotate a pentagon is 36 degrees
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I have this for my math review &amp; i just don't seem to get it...<br>what is (.9x+.8y)^2
elena55 [62]
Can you take the quantity [ x + y ] and square it ?

Can you take the quantity  [ x + 2y ] and square it ?

Then you should have no trouble with the square of (.9x + .8y) .

The square of any binomial (expression with 2 terms) is

(square of the first term)
+ (square of the second term)
+ 2 times (product of the terms) .
8 0
3 years ago
Solve using the square root property.X^2=-12
Alla [95]

Answer:

x = ± 2i\sqrt{3}

Step-by-step explanation:

note that \sqrt{-1} = i

x² = - 12 ( take square root of both sides )

x = ± \sqrt{-12} = ± \sqrt{4(3)(-1)} = ± 2i\sqrt{3}

5 0
2 years ago
The cost of 28 pounds horse feed is 63.28 what is the per pound of horse feed
Bas_tet [7]
$2.26 pounds of horse feed
8 0
3 years ago
-PLeAse HeLP-
ExtremeBDS [4]

This question is tricky.

I think there would be B)24 people that would prefer Yellow :)


4 0
3 years ago
3 questions in 1
MrRissso [65]

Answer:

1. D. 20, 30, and 50

2. A. 86

3. B. 94

Step-by-step explanation:

1. To find the outliers of the data set, we need to determine the Q1, Q3, and IQR.

The Q1 is the middle data in the lower part (first 10 data values) of the data set (while the Q3 is the middle data of the upper part (the last 10 data values) the data set.

Since it is an even data set, therefore, we would look for the average of the 2 middle values in each half of the data set.

Thus:

Q1 = (85 + 87)/2 = 86

Q3 = (93 + 95)/2 = 94

IQR = Q3 - Q1 = 94 - 86

IQR = 8

Outliers in the data set are data values below the lower limit or above the upper limit.

Let's find the lower and upper limit.

Lower limit = Q1 - 1.5(IQR) = 86 - 1.5(8) = 74

The data values below the lower limit (74) are 20, 30, and 50

Let's see if we have any data value above the upper limit.

Upper limit = Q3 + 1.5(IQR) = 94 + 1.5(8) = 106

No data value is above 106.

Therefore, the only outliers of the data set are:

D. 20, 30, and 50

2. See explanation on how to we found the Q1 of the given data set as explained earlier in question 1 above.

Thus:

Q1 = (85 + 87)/2 = 86

3. Q3 = (93 + 95)/2 = 94

3 0
2 years ago
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