Answer:
the Pythagorean formula is a²+b²=c², which is a and b are the x and y distances between points and c is the hypotenuse, which is the distance.
Step-by-step explanation:
Answer:
![(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C](https://tex.z-dn.net/?f=%28%5Cfrac%7Bx%5E%7B2%7D-25%7D%7B2%7D%29ln%285%2Bx%29-%5Cfrac%7Bx%5E%7B2%7D%7D%7B4%7D%2B%5Cfrac%7B5x%7D%7B2%7D%2BC)
Step-by-step explanation:
Ok, so we start by setting the integral up. The integral we need to solve is:
![\int x ln(5+x)dx](https://tex.z-dn.net/?f=%5Cint%20x%20ln%285%2Bx%29dx)
so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:
U=5+x
du=dx
x=U-5
so when substituting the integral will look like this:
![\int (U-5) ln(U)dU](https://tex.z-dn.net/?f=%5Cint%20%28U-5%29%20ln%28U%29dU)
now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:
![\int (pq')=pq-\int qp'](https://tex.z-dn.net/?f=%5Cint%20%28pq%27%29%3Dpq-%5Cint%20qp%27)
so we must define p, q, p' and q':
p=ln U
![p'=\frac{1}{U}dU](https://tex.z-dn.net/?f=p%27%3D%5Cfrac%7B1%7D%7BU%7DdU)
![q=\frac{U^{2}}{2}-5U](https://tex.z-dn.net/?f=q%3D%5Cfrac%7BU%5E%7B2%7D%7D%7B2%7D-5U)
q'=U-5
and now we plug these into the formula:
![\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU](https://tex.z-dn.net/?f=%5Cint%20%28U-5%29lnUdU%3D%28%5Cfrac%7BU%5E%7B2%7D%7D%7B2%7D-5U%29lnU-%5Cint%20%5Cfrac%7B%5Cfrac%7BU%5E%7B2%7D%7D%7B2%7D-5U%7D%7BU%7DdU)
Which simplifies to:
![\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU](https://tex.z-dn.net/?f=%5Cint%20%28U-5%29lnUdU%3D%28%5Cfrac%7BU%5E%7B2%7D%7D%7B2%7D-5U%29lnU-%5Cint%20%28%5Cfrac%7BU%7D%7B2%7D-5%29dU)
Which solves to:
![\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C](https://tex.z-dn.net/?f=%5Cint%20%28U-5%29lnUdU%3D%28%5Cfrac%7BU%5E%7B2%7D%7D%7B2%7D-5U%29lnU-%5Cfrac%7BU%5E%7B2%7D%7D%7B4%7D%2B5U%2BC)
so we can substitute U back, so we get:
![\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C](https://tex.z-dn.net/?f=%5Cint%20xln%28x%2B5%29dU%3D%28%5Cfrac%7B%28x%2B5%29%5E%7B2%7D%7D%7B2%7D-5%28x%2B5%29%29ln%28x%2B5%29-%5Cfrac%7B%28x%2B5%29%5E%7B2%7D%7D%7B4%7D%2B5%28x%2B5%29%2BC)
and now we can simplify:
![\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C](https://tex.z-dn.net/?f=%5Cint%20xln%28x%2B5%29dU%3D%28%5Cfrac%7Bx%5E%7B2%7D%7D%7B2%7D%2B5x%2B%5Cfrac%7B25%7D%7B2%7D-25-5x%29ln%285%2Bx%29-%5Cfrac%7Bx%5E%7B2%7D%2B10x%2B25%7D%7B4%7D%2B25%2B5x%2BC)
![\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C](https://tex.z-dn.net/?f=%5Cint%20xln%28x%2B5%29dU%3D%28%5Cfrac%7Bx%5E%7B2%7D-25%7D%7B2%7D%29ln%285%2Bx%29-%5Cfrac%7Bx%5E%7B2%7D%7D%7B4%7D-%5Cfrac%7B5x%7D%7B2%7D-%5Cfrac%7B25%7D%7B4%7D%2B25%2B5x%2BC)
![\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C](https://tex.z-dn.net/?f=%5Cint%20xln%28x%2B5%29dU%3D%28%5Cfrac%7Bx%5E%7B2%7D-25%7D%7B2%7D%29ln%285%2Bx%29-%5Cfrac%7Bx%5E%7B2%7D%7D%7B4%7D%2B%5Cfrac%7B5x%7D%7B2%7D%2BC)
notice how all the constants were combined into one big constant C.
Answer:
Step-by-step explanation:
To find the HCF of 144 and 180
By using product of prime method
Firstly express 144 as a product of it prime and express 180 as a product of it prime
140=2×2×2×2×3×3
180=2×2×3×3×5
Common factor =2×2×3×3
36
in term of m
m=36
13m-3
To find m
Substitute for m when m=36
13(36)-3=
465
The formula for exponential growth is y = amount(1+r)^x
Where amount would be the starting value, r would be the percentage and x would be the length of time.
The listed formula that meets that is D. y = 35(1.35)^x
Answer:
One of the most prevalent applications of exponential functions involves growth and decay models. Exponential growth and decay show up in a host of natural applications. From population growth and continuously compounded interest to radioactive decay and Newton’s law of cooling, exponential functions are ubiquitous in nature. In this section, we examine exponential growth and decay in the context of some of these applications.