The change in amount of gasoline is -0.55
<h3>Rate of change</h3>
The rate of change is the ratio of change in the y variable to that of the x variable.
From the given table, we can use the following coordinate (1, -2.15) and (2, -2.70)
Substitute the coordinate into the formula to have:
![Rate \ of \ change = \frac{-2.70+2.15}{2-1}\\Rate \ of \ change = \frac{-0.55}{1}\\Rate \ of \ change =-0.55](https://tex.z-dn.net/?f=Rate%20%5C%20of%20%5C%20change%20%3D%20%5Cfrac%7B-2.70%2B2.15%7D%7B2-1%7D%5C%5CRate%20%5C%20of%20%5C%20change%20%3D%20%5Cfrac%7B-0.55%7D%7B1%7D%5C%5CRate%20%5C%20of%20%5C%20change%20%3D-0.55)
Hence the change in amount of gasoline is -0.55
Learn more on rate of change here:brainly.com/question/8728504
This is the concept of algebra. The simplified form of the expression will be:
(8w³-10w²y^5)/(8w³x³)
=[2w²(4w-5y^5)]/[2w²(4wx³)]
2w² will cancel out and we shall remain with:
(4w-5y^5)/(4wx³)
The simplified form of our original expression will therefore be:
(4w-5y^5)/(4wx³)
Answer:
CD = 27
Step-by-step explanation:
Given the triangles are similar then the ratios of corresponding sided are equal, that is
=
, substitute values
=
= 3 ( multiply both sides by 9 )
CD = 27
Answer:
![a = 2^\alpha.14 \ ,\alpha\geq 8\\b = 7^\delta.14 \ , \delta\geq 3](https://tex.z-dn.net/?f=a%20%3D%202%5E%5Calpha.14%20%5C%20%2C%5Calpha%5Cgeq%208%5C%5Cb%20%3D%207%5E%5Cdelta.14%20%5C%20%2C%20%5Cdelta%5Cgeq%203)
Step-by-step explanation:
As 2 and 7 are the only prime divisors of both
and
we know that both can be written as:
![a = 2^\alpha . 7^\beta \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\](https://tex.z-dn.net/?f=a%20%3D%202%5E%5Calpha%20.%207%5E%5Cbeta%20%5C%20%2C%20%5Calpha%20%2C%5Cbeta%5C%20%5Cin%20%5Cmathbb%7BN%7D_%7B0%7D%5C%5Cb%20%3D%202%5E%5Cgamma%20.%207%5E%5Cdelta%20%5C%20%2C%20%5Cgamma%20%2C%5Cdelta%5C%20%5Cin%20%5Cmathbb%7BN%7D_%7B0%7D%5C%5C)
Where
is the set of all natural numbers adding the zero (careful because this part is important as I'll explain next).
We also know that 14 divides both numbers and that is actually the greatest common divisor between them. So we can rewrite a and b as follows:
![a = 2^\alpha . 7^\beta.14 \ , \alpha ,\beta\ \in \mathbb{N}_{0}\\b = 2^\gamma . 7^\delta.14 \ , \gamma ,\delta\ \in \mathbb{N}_{0}\\](https://tex.z-dn.net/?f=a%20%3D%202%5E%5Calpha%20.%207%5E%5Cbeta.14%20%5C%20%2C%20%5Calpha%20%2C%5Cbeta%5C%20%5Cin%20%5Cmathbb%7BN%7D_%7B0%7D%5C%5Cb%20%3D%202%5E%5Cgamma%20.%207%5E%5Cdelta.14%20%5C%20%2C%20%5Cgamma%20%2C%5Cdelta%5C%20%5Cin%20%5Cmathbb%7BN%7D_%7B0%7D%5C%5C)
Why do I write them like this? Because this way is easier to observe that if
and
were both greater than zero, then 28 would divide both hence 14 wouldn't be their g.c.d.. Likewise, if
and
were both greater than zero, then 98 would divide both and once again, 14 wouldn't be their g.c.d.
So either of them has to be equal to zero. And then we have that
![a = 2^\alpha.14 \\b = 7^\delta.14](https://tex.z-dn.net/?f=a%20%3D%202%5E%5Calpha.14%20%5C%5Cb%20%3D%207%5E%5Cdelta.14)
All we have left to do is find the possible values for
and
so that
and that only happens if
and ![\delta\geq 3](https://tex.z-dn.net/?f=%5Cdelta%5Cgeq%203)