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disa [49]
3 years ago
10

Simplify the radical

Mathematics
1 answer:
prisoha [69]3 years ago
6 0

Answer:

2 sqrt 6n/3n^2

Step-by-step explanation:

expand, multiply the fractions, calculate, multiply, calculate

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Find the equilibrium point for the pair of demand and supply functions. Here q represents the
olya-2409 [2.1K]

The equilibrium point for the pair of demand and supply function is 100

We have been two linear function, one is linear supply function and other is linear demand function.

In general , linear supply function is given as:

                            Qs = x + yP

                       Where , Qs = quantity supplied

                                        x = quantity

                                        P = price

And linear demand function is given is :

                             Qd = x + yP

                      Where , Qs = quantity supplied

                                        x = quantity

                                        P = price

According to the question,

                   Linear supply function is q = 300 + 5x

          And  linear demand function is q = 4800 – 40x

To find the equilibrium point we will put two quantities equal, that is,

                                 Qs = Qd

                      300 + 5x = 4800 – 40x

                        5x + 40x = 4800 – 300

                          45x = 4500

                             x =  100

Hence the equilibrium point is 100

Learn more about equilibrium point here : brainly.com/question/1915798

#SPJ9

4 0
2 years ago
The function p(x) is an odd degree polynomial with a negative leading coefficient.
Aliun [14]
A because it’s the right answer
4 0
3 years ago
Loli-Land Candy Store sells a 12 ounce bag of 60 mints. What is the number of mints per ounce?
Vsevolod [243]
5 because you have to divide 60 by 12.
6 0
3 years ago
First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x
e-lub [12.9K]

Answer:

(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

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

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

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

\int (pq')=pq-\int qp'

so we must define p, q, p' and q':

p=ln U

p'=\frac{1}{U}dU

q=\frac{U^{2}}{2}-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

Which simplifies to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU

Which solves to:

\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C

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

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

\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

\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C

notice how all the constants were combined into one big constant C.

7 0
3 years ago
What is the equation of the line that passes through the point (6,-5)
nexus9112 [7]

Answer:

y= -3/2x + 4

Step-by-step explanation:

hope this helps :)

6 0
2 years ago
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