since we have the area of the front side, to get its volume we can simple get the product of the area and the length, let's firstly change the mixed fractions to improper fractions.
![\stackrel{mixed}{23\frac{2}{3}}\implies \cfrac{23\cdot 3+2}{3}\implies \stackrel{improper}{\cfrac{71}{3}} ~\hfill \stackrel{mixed}{4\frac{7}{8}}\implies \cfrac{4\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{39}{8}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{71}{3}\cdot \cfrac{39}{8}\implies \cfrac{71}{8}\cdot \cfrac{39}{3}\implies \cfrac{71}{8}\cdot 13\implies \cfrac{923}{8}\implies 115\frac{3}{8}~in^3](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B23%5Cfrac%7B2%7D%7B3%7D%7D%5Cimplies%20%5Ccfrac%7B23%5Ccdot%203%2B2%7D%7B3%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B71%7D%7B3%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B4%5Cfrac%7B7%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B4%5Ccdot%208%2B7%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B39%7D%7B8%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B71%7D%7B3%7D%5Ccdot%20%5Ccfrac%7B39%7D%7B8%7D%5Cimplies%20%5Ccfrac%7B71%7D%7B8%7D%5Ccdot%20%5Ccfrac%7B39%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B71%7D%7B8%7D%5Ccdot%2013%5Cimplies%20%5Ccfrac%7B923%7D%7B8%7D%5Cimplies%20115%5Cfrac%7B3%7D%7B8%7D~in%5E3)
The common difference if there is one is the constant difference that occurs between any term and the term before it.... in this case:
There is no common difference,
dx=18,20,16,18 the difference or velocity is not constant...
d2x=2, -4,2 the acceleration is not constant...
d3x=-6,6 the thrust is not constant
Now we might be tempted to say that:
d4x=12 and say that that is constant and we COULD make a quartic equation fit all the data points, but without further data points in the sequence there is no mathematical proof that the quartic equation would produce accurate data points outside of the range given...
And solving a system of five equations for five unknowns is tedious for such a problem...a^4+bx^3+cx^2+dx+e=y
Answer:
The co-ordinates of the vertex of the function y-9= -6(x-1)^2 is (1, 9)
<u>Solution:</u>
Given, equation is 
We have to find the vertex of the given equation.
When we observe the equation, it is a parabolic equation,
We know that, general form of a parabolic equation is
Where, h and k are x, y co ordinates of the vertex of the parabola.
By comparing the above equation with general form of the parabola, we can conclude that,
a = -6, h = 1 and k = 9
Hence, the vertex of the parabola is (1, 9).
Answer:
1. B. point B.
2. C. point C.
3. C. point C.
Step-by-step explanation:
1. In order to find the graph's y-intercept, we need to locate the point where the line crosses the y-axis. This will always happen at x=0, therefore, the y-intercept is located at point B (0,-4)
2. In order to find the x-intercept, we need to find the point where the line crosses the x-axis. This will generally happen when y=0, so that will be point C (2,0)
3. In order to find the graph's zero, we need to find the point where y=0. In other words, the graph's zero is the point where the function is equal to zero (the x-intercept) so this will br point C again (2,0)
