Answer:
no solutions
Step-by-step explanation:
5 -2 sqrt(x) = 11
Subtract 5 from each side
5- 5 -2 sqrt(x) = 11-5
-2 sqrt (x) = 6
Divide each side by -2
-2/-2 sqrt(x) = 6/-2
sqrt(x) = -3
The sqrt(x) must be 0 or positive
no solutions
Answer:
167,886,383 basketballs
Step-by-step explanation:
We are asked to find how many basketballs would take to circle around the equator. We have given the earth's radius. So, we need the formula to obtain it's perimeter. Thus:
![Perimeter = Pi*diameter [m]\\Where Pi = 3.14](https://tex.z-dn.net/?f=Perimeter%20%3D%20Pi%2Adiameter%20%5Bm%5D%5C%5CWhere%20Pi%20%3D%203.14)
Earth's diameter is simply radius*2. It means:
![P= 3.14*6400*2 = 40,192 [km]](https://tex.z-dn.net/?f=P%3D%203.14%2A6400%2A2%20%3D%2040%2C192%20%5Bkm%5D)
On the other hand, we have a basketball crcumference; however, we need to obtain its diameter so that we can later calculate how many basketballs fit on earth's equator by simply dividing earth's circumference by a basketball's diameter.
Diameter of a basketball:
![D= Perimeter /Pi [m]](https://tex.z-dn.net/?f=D%3D%20Perimeter%20%2FPi%20%5Bm%5D)
We need to change units to fit in the international system.
29.6 in to cm = ![29.6*2.54 = 75.184 [cm]](https://tex.z-dn.net/?f=29.6%2A2.54%20%3D%2075.184%20%5Bcm%5D)
Then:
![Diameter=75.184/3.14 = 23.94 [cm]](https://tex.z-dn.net/?f=Diameter%3D75.184%2F3.14%20%3D%2023.94%20%5Bcm%5D)
We have to convert earth's perimeter in km to cm:
![Equator=40192[km]*100000=4,019,200,000 [cm]](https://tex.z-dn.net/?f=Equator%3D40192%5Bkm%5D%2A100000%3D4%2C019%2C200%2C000%20%5Bcm%5D)
Finally, dividing total earth's circumference by a basketball diameter:
![Totalbasketballs=4,019,200,000/23.94= 167,886,383 [basketballs]](https://tex.z-dn.net/?f=Totalbasketballs%3D4%2C019%2C200%2C000%2F23.94%3D%20167%2C886%2C383%20%5Bbasketballs%5D)
Answer:
63.56 -81.47= -17.91
Step-by-step explanation:
cause the answer of 63.56 +(-81.47)= -17.91
Number 1 & 2 both goes in 200. Numbers 9,3, & 27 goes into 81. Numbers 10,5,4,8,20, & 25 goes into 200. Hope this helps
Ax^2 + bx + c
An expression with a quadratic structure will have a term being raised to the second power