The benchmarks are: 0, 0.25, 0.50, 0.75 and 1.
2. 81 → 2.75
+
3.73 → 3.75
------------
2.75 + 3.75 = 6.50
When finding the domain of a square root, you have to know that it is impossible to get the square root of 0 or any negative number. since domain is possible x values this means that x cannot be 0 or any number less than 0. However, you can find the square root of the smallest most infinitely small number greater than 0. since an infinitely small number close to zero can not be written out, we must must say that the domain starts at 0 exclusive. exclusive is represented by an open or close parenthesis so in this case the domain starts with:
(0,
we can get the square root of any number larger than 0 up to infinity but infinity can never be reached so it is also exclusive. So so the ending of our domain would be:
,infinity)
So the answer if the square root is only over the x the answer is
(0, infinity)
But if the square root is over the x- 5 then this would brIng a smaller amount of possible x values. since anything under the square root sign has to be greater than 0, you can say that:
(x - 5) > 0
x > 5
Therefore the domain would start at 5 and the answer would be:
(5, infinity)
Answer: A hypothesis is like a prediction
Hope it helps
Answer:
Point D
Step-by-step explanation:
AD and CD are a line
The intersection of two lines is a point
The two lines intersect at point D
Answer:
![DG = 26](https://tex.z-dn.net/?f=DG%20%3D%2026)
![GE = 26](https://tex.z-dn.net/?f=GE%20%3D%2026)
![DF = 15](https://tex.z-dn.net/?f=DF%20%3D%2015)
![CH = 14](https://tex.z-dn.net/?f=CH%20%3D%2014)
![CE = 28](https://tex.z-dn.net/?f=CE%20%3D%2028)
Step-by-step explanation:
The figure has been attached, to complement the question.
![DE = 52](https://tex.z-dn.net/?f=DE%20%3D%2052)
![FC = 15](https://tex.z-dn.net/?f=FC%20%3D%2015)
![HE = 14](https://tex.z-dn.net/?f=HE%20%3D%2014)
Given that J is the centroid, it means that J divides sides CD, DE and CE into two equal parts respectively and as such the following relationship exist:
![DF = FC](https://tex.z-dn.net/?f=DF%20%3D%20FC)
![CH = HE](https://tex.z-dn.net/?f=CH%20%3D%20HE)
![DG = GE](https://tex.z-dn.net/?f=DG%20%3D%20GE)
Solving (a): DG
If
, then
![DE = DG + GE](https://tex.z-dn.net/?f=DE%20%3D%20DG%20%2B%20GE)
![DE = DG + DG](https://tex.z-dn.net/?f=DE%20%3D%20DG%20%2B%20DG)
![DE = 2DG](https://tex.z-dn.net/?f=DE%20%3D%202DG)
Make DG the subject
![DG = \frac{1}{2}DE](https://tex.z-dn.net/?f=DG%20%3D%20%5Cfrac%7B1%7D%7B2%7DDE)
Substitute 52 for DE
![DG = \frac{1}{2} * 52](https://tex.z-dn.net/?f=DG%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%2052)
![DG = 26](https://tex.z-dn.net/?f=DG%20%3D%2026)
Solving (b): GE
If
, then
![GE = DG](https://tex.z-dn.net/?f=GE%20%3D%20DG)
![GE = 26](https://tex.z-dn.net/?f=GE%20%3D%2026)
Solving (c): DF
![DF = FC](https://tex.z-dn.net/?f=DF%20%3D%20FC)
So:
![DF = 15](https://tex.z-dn.net/?f=DF%20%3D%2015)
Solving (d): CH
![CH = HE](https://tex.z-dn.net/?f=CH%20%3D%20HE)
![CH = 14](https://tex.z-dn.net/?f=CH%20%3D%2014)
Solving (e): CE
If
, then
![CE = CH + HE](https://tex.z-dn.net/?f=CE%20%3D%20CH%20%2B%20HE)
![CE = 14 + 14](https://tex.z-dn.net/?f=CE%20%3D%2014%20%2B%2014)
![CE = 28](https://tex.z-dn.net/?f=CE%20%3D%2028)