2(4x - 9) =
2 * 4x - 2 * 9 =
8x - 18
Hope this helps!!
~Kiwi
How do you become an administrator?
I think answering a lot of questions and dont get warnings
The distance (d) between two points (x1,y1) and (x2,y2) is given by the formula
d = √ ((X2-X1)2+(Y2-Y1)2)
d = √ (-400--800)2+(300-200)2
d = √ ((400)2+(100)2)
d = √ (160000+10000)
d = √ 170000
The distance between the points is 412.310562561766
The midpoint of two points is given by the formula
Midpoint= ((X1+X2)/2,(Y1+Y2)/2)
Find the x value of the midpoint
Xm=(X1+X2)/2
Xm=(-800+-400)/2=-600
Find the Y value of the midpoint
Ym=(Y1+Y2)/2
Ym=(200+300)/2=250
The midpoint is: (-600,250)
Graphing the two points, midpoint and distance
P1 (-800,200)
P2 (-400,300)
Midpoint (-600,250)
The length of the black line is the distance between the points (412.310562561766)
Hello!
I would be more than happy to help!
So to find the balance use this equation.
Balance = Credit- Debit
In other words, to find BALANCE... You Subtract Debit from Credit
whats my credit? 84$
Whats my debit? 29$
Whats my balance? Lets find out! How? Well, lets subtract my credit (84) from my debit (29).
84-29= 55
Balance =55
Hope this helped!
Answer:
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola
y=5−x^2. What are the dimensions of such a rectangle with the greatest possible area?
Width =
Height =
Width =√10 and Height 
Step-by-step explanation:
Let the coordinates of the vertices of the rectangle which lie on the given parabola y = 5 - x² ........ (1)
are (h,k) and (-h,k).
Hence, the area of the rectangle will be (h + h) × k
Therefore, A = h²k ..... (2).
Now, from equation (1) we can write k = 5 - h² ....... (3)
So, from equation (2), we can write
![A =h^{2} [5-h^{2} ]=5h^{2} -h^{4}](https://tex.z-dn.net/?f=A%20%3Dh%5E%7B2%7D%20%5B5-h%5E%7B2%7D%20%5D%3D5h%5E%7B2%7D%20-h%5E%7B4%7D)
For, A to be greatest ,
⇒ ![h[10-4h^{2} ]=0](https://tex.z-dn.net/?f=h%5B10-4h%5E%7B2%7D%20%5D%3D0)
⇒ 
⇒ 
Therefore, from equation (3), k = 5 - h²
⇒ 
Hence,
Width = 2h =√10 and
Height = 
