X=y-4 Let's make this eqn 1 and
2x-5y=3 this eqn 2..
Two things are possible here...
Its either you substitute for x in equation 2... Or you substitute for y...
x=y-4 put this in the equation 2..
2x-5y=3.. Replace the x with (y-4)
2(y-4)-5y = 3...
Since this is In the option no point in substituting for y...
Hope this helped...?
Answer:
0.83 miles per minute
Step-by-step explanation:
Please kindly check the attached files for details.
Answer: E. y(x) = 0
Step-by-step explanation:
y(x) = 0 is the only answer from the options that satisfies the differential equal y" - 4y' + 4y = 0
See:
Suppose y = e^(-2x)
Differentiate y once to have
y' = -2e^(-2x)
Differentiate the 2nd time to have
y" = 4e^(-2x)
Now substitute the values of y, y', and y" into the give differential equation, we have
4e^(-2x) - 4[-2e^(-2x)] + 4e^(-2x)
= 4e^(-2x) + 8e^(-2x) + 4e^(-2x)
= 16e^(-2x)
≠ 0
Whereas we need a solution that makes the differential equation to be equal to 0.
If you test for the remaining results, the only one that gives 0 is 0 itself, and that makes it the only possible solution from the options.
It is worth mentioning that apart from the trivial solution, 0, there is a nontrivial solution, but isn't required here.
22a is < since multiplying something by a fraction less than one makes it smaller.
22b is = since multiplying something by a fraction equal to one makes it stay the same.
22c is > since multiplying something by a mixed number makes it larger.
22d is < since multiplying something by a fraction less than one makes it smaller.
Hope this helps!
Answer:
114 kg
Step-by-step explanation:
<u>Step 1: Make an expression</u>
Decrease 120 kg by 5%
<u>Step 2: Multiply</u>
Answer: 114 kg
