Let x,y be two different numbers
suppose x^2=y^2
then x^2-y^2=0
which yields (x+y)(x-y)=0
so either x=y or x=-y
In any case, x and y must be the same value
also when a vairable is squared like y=x^2
we must note that there are 2 possible solutions
x=(+/-)sqrt(y)
Answer:
Step-by-step explanation:
a. The five-number summary is made up of the following summary means:
1. Minimum: 25
2. First Quartile: 30.5
3. Medium: 34.5
4. Third quartile: 43
5. Maximum: 78
6. P30: 32.1
b.
Standard Deviation: 12.72
Rank: 53
Interquartile range: 12.5
C. The interquartile range is 12.5 and 1.5 times the interquartile range is (1.5) (12.5) = 18.75. Third quartile plus 1.5 times the interquartile range is 61.75. The value of 78 exceeds 61.75, then 78 is an outlier.
x = 4Rearrange:Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
1/8*x-(1/2)=0
Step by step solution :Step 1 : 1 Simplify — 2Equation at the end of step 1 : 1 1 (— • x) - — = 0 8 2Step 2 : 1 Simplify — 8Equation at the end of step 2 : 1 1 (— • x) - — = 0 8 2Step 3 :Calculating the Least Common Multiple : 3.1 Find the Least Common Multiple
The left denominator is : 8
The right denominator is : 2
the answer is 4
Volume =
(pi)(
(height)
$Cost = ($0.09)(3.14)(9*9)(9.5) = $217.4607
1. C
2.D
3.A
4.B
Hope this helps!
