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2 years ago

What is the value of the expression? 9X3-9x 20 -5

Mathematics

1 answer:

Tom [10]2 years ago

3 0

Answer:

9x^3-5

Step-by-step explanation:

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Ricardo took the following steps to simplify the expression below but was surprised to hear that his answer was incorrect. Which

Xelga [282]

You have not given us any of the steps that Ricardo took to simplify the
expression, and you also haven't given us the list of choices that includes
the description of his mistake, so you're batting O for two so far.
Other than those minor details, the question is intriguing, and it certainly
draws me in.

If Ricardo made a mistake in simplifying that expression, I'm going to say that
it was most likely in the process of removing the parentheses in the middle.

Now you understand that this is all guess-work, because of all the stuff that you
left out when you copied the question, but I think he probably forgot that the 3x
operates on everything inside the parentheses.

He probably wrote that 3x (x-3) is

either 3x² - 3
or x - 9x .

In reality, when properly simplified,

3x (x - 3) = 3x² - 9x .

3 0

2 years ago

Write a number sentence for the problem. Usefor the missing number. Then solve . 29 children rode their bikes to school. After s

Elza [17]

Answer:

well apperently all will ride home but i need you to type the question correctly so i know what your asking. :)

Step-by-step explanation:

6 0

2 years ago

Tell whether c = 3 is a solution of c + 7 = 10.<br> a) Yes<br> b) No

Svet_ta [14]

Answer:

a) Yes

Step-by-step explanation:

Because, 3+7=10

4 0

2 years ago

Read 2 more answers

How do you write (10, -6) and (-2, -6) as an equation

Paraphin [41]

Answer:

y=-6

Step-by-step explanation:

Method 1:

Both the y coordinates of the points are the same and the x coordinates are distinct.

Therefore the straight line passing through the point is parallel to X-axis.

The common y coordinate is -6, therefore the line equation is y=-6.

Method 2:

We will solve it using general line equation

(y-y1)=( $\frac{y2-y1}{x2-x1}$ )(x-x1)

where (x1,y1) and (x2,y2) are the 2 points.

(y+6)=( $\frac{-6+6}{10+2}$ )(x+2)

(y+6)=0

y=-6

4 0

3 years ago

<img src="https://tex.z-dn.net/?f=%7B%5Chuge%7B%5Cunderline%7B%5Csmall%7B%5Cmathbb%7B%5Cpink%7BREFER%20%5C%20TO%20%5C%20THE%20%5

marusya05 [52]

Answer:

See below

(B) and (C) are correct.

Step-by-step explanation:

We have the following limit

$\lim \limits_{n\rightarrow \infty} \left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} }, \forall x>0$

I am not sure about methods concerning the quotient, but in this type of question I would try to convert this limit into integration.

Considering the numerator, we have

$(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right) = \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)$

- I didn't forget about $n^n$

Considering the denominator, we have

$(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right) = \prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)$

- I didn't forget about $n!$

Therefore,

$\left(\frac{n^n(x+n) \left(x+\dfrac{n}{2} \right)\dots \left(x+\dfrac{n}{n} \right)}{n!(x^2+n^2)\left(x^2+\dfrac{n^2}{4} \right)\dots \left(x^2+\dfrac{n^2}{n^2} \right)}\right)^{\dfrac{x}{n} } = \left(\dfrac{n^2 \prod_{k=1}^n \left(x+\dfrac{n}{k} \right)}{ n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right)} \right)$

$= \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

Now we have

$\lim \limits_{n\rightarrow \infty} \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}, \forall x>0$

This is just the notation change so far.

What I want to do here is apply definite integrals using Riemann Integrals (We will write the limit as an definite integral). A nice way to do it is using logarithms. Therefore, we can apply the natural logarithm in both sides.

Now, recall two properties of logarithms:

$\boxed{\log_a mn = \log_a m + \log_a n}$

$\boxed{\log_a m^p = p\log_a m}$

$\boxed{\log_a \left(\dfrac{m}{n} \right) = \log_a m- \log_a n}$

Thus,

$\ln f(x) = \lim \limits_{n\rightarrow \infty} \ln \left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)^{\dfrac{x}{n}}$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n}\ln\left(\dfrac{n^n}{n!}\prod_{k=1}^n\dfrac{\left(x+\dfrac{n}{k}\right)}{\left(x^2+\dfrac{n^2}{k^2}\right)}\right)$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln \left(n^n \prod_{k=1}^n \left(x+\dfrac{n}{k} \right) \right)-\ln \left( n!\prod_{k=1}^n \left(x^2+\dfrac{n^2}{k^2} \right) \right) \right]$

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[\ln n^n + \prod_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)-\ln n! -\prod_{k=1}^n \ln\left(x^2+\dfrac{n^2}{k^2} \right) \right]$

Considering

$\lim \limits_{n\rightarrow \infty} \frac{x}{n} (\ln n^n - \ln n! ) = \lim \limits_{n\rightarrow \infty} \frac{x}{n} (n\ln n - \ln n! )= \lim \limits_{n\rightarrow \infty} \frac{x \cdot\ln\frac{n^n}{n!} }{n}$

Using Stirling's formula

$\dfrac{n^n}{n!}\underset{\infty}{\sim} \dfrac{n^n}{\sqrt{2n \pi}\left(\dfrac{n}{e}\right)^n}=\dfrac{e^n}{\sqrt{2n \pi}}$

then

$\ln\left(\frac{n^n}{n!}\right)\underset{\infty}{=}n\ln\left(e\right)-\frac{1}{2}\ln\left(2n\pi\right)+o\left(1\right)$

$\implies \frac{\ln\left(\frac{n^n}{n!}\right)}{n}=1-\frac{\ln(2n\pi)}{2n}+o\left(1\right)$

This shows our limit equals 1 as $\frac{\log(2\pi n)}{2n} \rightarrow 0$ and $\ln(e)=1$

Employing a Riemann sum in the main limit, we have

$= \lim \limits_{n\rightarrow \infty} \dfrac{x}{n} \left[ \sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) - \sum_{k=1}^n\ln \left(x^2+\dfrac{n^2}{k^2} \right) \right]$