So you have 85 pages and 10 days.
You would do 85/10 to get 8.5 pages each day. Hope that’s right !
B 2/10 is not simplified because you can divide each number by 2 reducing it to 1/5.
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 ![= \frac{10}{4}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B10%7D%7B4%7D)
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 ,
![\frac{dA}{dh} =0 = 10h-4h^{3}](https://tex.z-dn.net/?f=%5Cfrac%7BdA%7D%7Bdh%7D%20%3D0%20%3D%2010h-4h%5E%7B3%7D)
⇒ ![h[10-4h^{2} ]=0](https://tex.z-dn.net/?f=h%5B10-4h%5E%7B2%7D%20%5D%3D0)
⇒ ![h^{2} =\frac{10}{4} {Since, h≠ 0}](https://tex.z-dn.net/?f=h%5E%7B2%7D%20%3D%5Cfrac%7B10%7D%7B4%7D%20%7BSince%2C%20h%E2%89%A0%200%7D)
⇒ ![h = ±\frac{\sqrt{10} }{2}](https://tex.z-dn.net/?f=h%20%3D%20%C2%B1%5Cfrac%7B%5Csqrt%7B10%7D%20%7D%7B2%7D)
Therefore, from equation (3), k = 5 - h²
⇒ ![k=5-\frac{10}{4} =\frac{10}{4}](https://tex.z-dn.net/?f=k%3D5-%5Cfrac%7B10%7D%7B4%7D%20%3D%5Cfrac%7B10%7D%7B4%7D)
Hence,
Width = 2h =√10 and
Height = ![k =\frac{10}{4}.](https://tex.z-dn.net/?f=k%20%3D%5Cfrac%7B10%7D%7B4%7D.)
Answer:
6(base) x 3.5(Height) = 21, 21 divided by 2 (x 1/2) = 10.5
10.5(area of triangle) x 7.75(width) = 81.375
Formula of an area of a triangle:
BH x 1/2
Base x Height x 1/2(basically just dividing by 2)
Now the base:
1.5 x 7.75 x 6 = 69.75
Now let's add the two given volumes:
81.375+ 69.75 = 151.125 ft^3 is your answer.
![\bf 1:1.618\iff \cfrac{1}{1.618}\cfrac{\leftarrow \textit{small side=width}}{\leftarrow \textit{longer side=length}}\\\\ -----------------------------\\\\ \textit{now, what if the length is 35?}\to \cfrac{1}{1.618}=\cfrac{w}{\boxed{35}}](https://tex.z-dn.net/?f=%5Cbf%201%3A1.618%5Ciff%20%5Ccfrac%7B1%7D%7B1.618%7D%5Ccfrac%7B%5Cleftarrow%20%5Ctextit%7Bsmall%20side%3Dwidth%7D%7D%7B%5Cleftarrow%20%5Ctextit%7Blonger%20side%3Dlength%7D%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bnow%2C%20what%20if%20the%20length%20is%2035%3F%7D%5Cto%20%5Ccfrac%7B1%7D%7B1.618%7D%3D%5Ccfrac%7Bw%7D%7B%5Cboxed%7B35%7D%7D)
solve for "w" to get the width
what's the area of a rectangle? length * width, or lw