For this sort of problem, you need to be familiar with prefixes.
So:
megabyte = 
bytes
kilobyte = 
bytes
and a regular byte would just be 
Now, you will need to do a conversion:
I'm going to explain this just in case, but we convert megabytes to regular bytes in the first half by using the information I gave you above. (In one megabyte, there are 
bytes!)
In the second fraction thought, remember one kilobyte is bytes. However, you usually only see one of each thing on the bottom of fractions, so you need to add a negative sign to the 3. (technically you could just do 
, but I think it is more correct to do write it out how I did).
*** They do equal the same thing though! Do whichever way is easier for you!
Now, your answer should be:
kilobytes (1000 kilobytes)
Hope this helped!
6* 25 = 150 = (6*20= 120) + (6*5=30) Then Add 120+30= 150 so your answer is 5
Answer: C is correct, No solution
Step-by-step explanation:
Let’s simplify the second equation,
8x-4y=-20
We can divide the whole equation by 4...
8x/4-4y/4=-20/4
Which becomes....
2x-y=-5
Now let’s move things around to see if it’s the same equation as the one on top...
-y=-2x-5
y=2x+5
Now we need to solve the systems...
y=2x-5
y=2x+5
Since their slopes are both 2 and their y-intercepts are not identical, the answer is no solution. The two lines will continue on without ever crossing because they have the same slope.
Answer with explanation:
--------------------------------------------------------Dividing both sides by 8 x
This Integration is of the form ⇒y'+p y=q,which is Linear differential equation.
Integrating Factor
Multiplying both sides by Integrating Factor
![x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\times [y'+y\times\frac{1+4x^2}{8x}]=\frac{1}{8}\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}\\\\ \text{Integrating both sides}\\\\y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\frac{1}{8}\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx \\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=\int {x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}} \, dx\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=-[x^{\frac{9}{8}}]\times\frac{ \Gamma(0.5625, -x^2)}{(-x^2)^{\frac{9}{16}}}\\\\8y\times x^{\frac{1}{8}}\times e^{\frac{x^2}{2}}=(-1)^{\frac{-1}{8}}[ \Gamma(0.5625, -x^2)]+C-----(1)](https://tex.z-dn.net/?f=x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5Ctimes%20%5By%27%2By%5Ctimes%5Cfrac%7B1%2B4x%5E2%7D%7B8x%7D%5D%3D%5Cfrac%7B1%7D%7B8%7D%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%5C%5C%5C%5C%20%5Ctext%7BIntegrating%20both%20sides%7D%5C%5C%5C%5Cy%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%20%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%5Cint%20%7Bx%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%7D%20%5C%2C%20dx%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D-%5Bx%5E%7B%5Cfrac%7B9%7D%7B8%7D%7D%5D%5Ctimes%5Cfrac%7B%20%5CGamma%280.5625%2C%20-x%5E2%29%7D%7B%28-x%5E2%29%5E%7B%5Cfrac%7B9%7D%7B16%7D%7D%7D%5C%5C%5C%5C8y%5Ctimes%20x%5E%7B%5Cfrac%7B1%7D%7B8%7D%7D%5Ctimes%20e%5E%7B%5Cfrac%7Bx%5E2%7D%7B2%7D%7D%3D%28-1%29%5E%7B%5Cfrac%7B-1%7D%7B8%7D%7D%5B%20%5CGamma%280.5625%2C%20-x%5E2%29%5D%2BC-----%281%29)
When , x=1, gives , y=9.
Evaluate the value of C and substitute in the equation 1.
true
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