If you are solving for "z", the answer is
z=2w^2+7w/10 -2
Answer:
x = 4
Step-by-step explanation:
Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :
-5*x+15-(35-10*x)=0
Pull out like factors :
5x - 20 = 5 • (x - 4)
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Solve : 5 = 0
This equation has no solution.
A a non-zero constant never equals zero.
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Solve : x-4 = 0
Add 4 to both sides of the equation :
x = 4
Answer:
Option 4
Step-by-step explanation:
The volume of a cylinder is going to be the area of the base (which is a circle) times the height. if you have cylindrical mug, then you can use the base area×height relationship, but it won't be exact because some of the volume is going to be used for the walls of the mug itself.
Anyways:
Volume = Base Area × Height
•Base Area -> Circle = π×radius²
•Height is given as 7in.
Volume = π×(6.2in)² × 7in = 845.33975...
Volume is in units cubed (the 3rd power) which is where the in³ is from. Your closest option here looks to be the 4th choice.
So... let's say the smaller regular octagon has sides of "x" long, then the larger octagon will have sides of 5x.
![\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &Sides&Area&Volume\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array} \\\\ -----------------------------\\\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7Bratio%20relations%7D%0A%5C%5C%5C%5C%0A%5Cbegin%7Barray%7D%7Bccccllll%7D%0A%26Sides%26Area%26Volume%5C%5C%0A%26-----%26-----%26-----%5C%5C%0A%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%26%5Ccfrac%7Bs%7D%7Bs%7D%26%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%26%5Ccfrac%7Bs%5E3%7D%7Bs%5E3%7D%0A%5Cend%7Barray%7D%20%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C)
We have been given the sequence 2,3,5,9,17.
We can write the terms of this sequence as
From the above term we can see that for the first term we take exponent 0 on 2 and then add 1 .
For second term we take exponent 1 on 2 and then add 1 .
For third term we take exponent 2 on 2 and then add 1 .
Using this fact for the next term of the sequence i.e. 6th term, we can take exponent 5 on 2 and then add 1 .
Therefore, next term of the sequence is given by
Therefore, the next term is 33.
Using the above facts, the pattern is given by
