G(x) is a quadratic function with a parabolic graph that opens up and has vertex at (0,0). If we dilate this vertically by a factor of 6, we get h(x) = 6x^2.
If we translate the graph of h(x) 1 unit up and 4 units to the right, we get
k(x) = 6(x-4)^2 + 1.
The solution depends on the value of

. To make things simple, assume

. The homogeneous part of the equation is

and has characteristic equation

which admits the characteristic solution

.
For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be

. Then

So you have


This means


and so the general solution would be
Answer:
19.4 liters
Step-by-step explanation:
You get 19.4 by multiplying 9.7 by 2 which = 19.4
Answer:
When the volume of a gas in a container varies inversely with the pressure on the gas and a container of nitrogen has a volume of 29.5 litres with 2000 psi. So, if the tank has a 14.7 psi pressure, then the volume of the tank would be 4013.6 litres.
Step-by-step explanation:
- The volume of a gas in a container varies inversely with the pressure on the gas. If a container of nitrogen has a volume of litres with psi, what is the volume if the tank has a psi pressure.
- Calculate the volume to the nearest whole number.
- Use the given values and calculate the proportionality constant as well as required solution.
Step 1 of 2
Write the inversely proportionality relation between the volume and pressure of a gas in container.

Where volume is represented by v and pressure is represented by p.
Remove the proportionality and place a constant say k.

Substitute the given values of nitrogen that v=29.5 and
p=2000 and calculate the value of k from above equation.

Step 2 of 2
Substitute the value p=14.7 and above calculated k value in the equation

Answer:
1. Rewriting the expression 5.a.b.b.5.c.a.b.5.b using exponents we get: 
5. 
6. 
7. 
Step-by-step explanation:
Question 1:
We need to rewrite the expression using exponents
5.a.b.b.5.c.a.b.5.b
We will first combine the like terms
5.5.5.a.a.b.b.b.b.c
Now, if we have 5.5.5 we can write it in exponent as: 
a.a as
b.b.b.b as: 
So, our result will be:

Rewriting the expression 5.a.b.b.5.c.a.b.5.b using exponents we get: 
Question:
Rewrite using positive exponent:
The rule used here will be:
which states that if we need to make exponent positive, we will take it to the denominator.
Applying thee above rule for getting the answers:
5)
6) 
7) 
We know that
so, we get
