All categories

3 years ago

Find the product. (3p+2)(5p-7)

Mathematics

2 answers:

ryzh [129]3 years ago

6 0

15p^2-11p-14 is the product

g100num [7]3 years ago

5 0

5p X -29p = - 10 this is your answer

You might be interested in

Please calculate this limit please help me

Tasya [4]

Answer:

We want to find:

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n}$

Here we can use Stirling's approximation, which says that for large values of n, we get:

$n! = \sqrt{2*\pi*n} *(\frac{n}{e} )^n$

Because here we are taking the limit when n tends to infinity, we can use this approximation.

Then we get.

$\lim_{n \to \infty} \frac{\sqrt[n]{n!} }{n} = \lim_{n \to \infty} \frac{\sqrt[n]{\sqrt{2*\pi*n} *(\frac{n}{e} )^n} }{n} = \lim_{n \to \infty} \frac{n}{e*n} *\sqrt[2*n]{2*\pi*n}$

Now we can just simplify this, so we get:

$\lim_{n \to \infty} \frac{1}{e} *\sqrt[2*n]{2*\pi*n} \\$

And we can rewrite it as:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n}$

The important part here is the exponent, as n tends to infinite, the exponent tends to zero.

Thus:

$\lim_{n \to \infty} \frac{1}{e} *(2*\pi*n)^{1/2n} = \frac{1}{e}*1 = \frac{1}{e}$

7 0

3 years ago

2(g+5)=-7-4g+g include instructions plz!

Elza [17]

First distribute the 2, then add 7 to that side, so you have 2g+17=-4g+g , subtract 2g and it should be easy to find from there lol

4 0

3 years ago

As a salesperson, Jonathan is paid $50 per week plus 3% of the total amount he sells. This week, he wants to earn at least $100.

Hoochie [10]

Answer:

sorry ima need to use your page

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

X^2+16x+64 can anyone show the work for this and find the factor.

Alexandra [31]

Answer:

should i simplify it? or factorise it?

I guess its factorise! So i m factorising!

x^2+16+64

(x)^2+2×x×8+(8)^2

(x+8)^2

=(x+8)(x+8)

8 0

3 years ago

What is the solution of the equation sqrt 2 x+4-sqrt x=2

ivann1987 [24]

There are 2 answers to this

x=0
x=16

3 0

3 years ago