All categories

2 years ago

Four less than the product of 8 and a number equals 6

Mathematics

1 answer:

docker41 [41]2 years ago

8 0

The equation will be 6x - 4= 8.

Solve the equation: 6x - 4 = 8 + 4 + 4.

6x = 12. 6 6. x = 2

Hopefully, that helps

You might be interested in

An evergreen nursery usually sells a certain shrub after 7 years of growth and shaping. The growth rate during those 7 years is

neonofarm [45]

Answer:

(a) $h(t)=\frac{1.3t^2}{2} + 2t +17$

(b)62.85cm

Step-by-step explanation:

The growth rate of the shrub is given as:

$\frac{dh}{dt} = 1.3t + 2$

Where t=time in years, h=height in centimeters

(a)First, we solve for the height h(t) by integrating.

$\int \frac{dh}{dt} dt= \int (1.3t + 2)dt\\h(t)=\frac{1.3t^2}{2} + 2t +C, $ C a constant of Integration$\\$We sunstitute the initial value to find the value of C$\\$When t=0, h=17cm$\\17=C\\Therefore:\\h(t)=\frac{1.3t^2}{2} + 2t +17$

(b)The shrub are sold after 7 years of growth. Therefore, we determine the value of h(t) when t=7 years.

$h(t)=\frac{1.3t^2}{2} + 2t +17\\h(7)=\frac{1.3*7^2}{2} + 2(7) +17\\=31.85+14+17\\h(7)=62.85cm$

The Shrubs are 62.85cm tall when they are sold.

4 0

2 years ago

If g(x) = 2x - 5 and h(x) = 3x + 7, then g(h(x)) = ?

lana [24]

Step-by-step explanation:

$g(x) = 2x - 5$

$h(x) = 3x + 7$

To determine $g(h(x))$ , wherever you see an $x$ in $g(x)$ , replace it with the equation $h(x)$ :

$g(x) = 2x - 5$

$g(h(x)) = 2(h(x)) - 5$

$g(h(x)) = 2(3x + 7) - 5$

$g(h(x)) = 6x + 14 - 5$

$g(h(x)) = 6x + 9$

8 0

2 years ago

You have $85 in your pocket. Each week you plan to save $8 from your allowance and $15 from your paycheck. How many weeks from n

Ede4ka [16]

At least 5 weeks. maybe

3 0

3 years ago

Read 2 more answers

<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

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

Now dividing the terms inside the parenthesis by $\dfrac{n}{k}$ in $\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right)$

we have

$\sum_{k=1}^n \ln \left(x+\dfrac{n}{k} \right) = \sum_{k=1}^n \ln \left(\frac{kx}{n} +1\right)$

Now dividing the terms inside the parenthesis by $\dfrac{n^2}{k^2}$ in $\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right)$

we have

$\sum_{k=1}^n \ln \left(x^2+\dfrac{n^2}{k^2} \right) = \sum_{k=1}^n \ln \left(\frac{(kx)^2}{n^2} +1\right)$

Therefore

$= \frac xn\sum_{k=1}^n\ln\dfrac{z+1}{z^2+1}$

for $\dfrac{kx}{n} = z$

Using Riemann Integral,

$\lim \limits_{n\rightarrow \infty} \int_0^1\ln\frac{z+1}{z^2+1}dz$

From

$\frac{f'(x)}{f(x)}=\ln\frac{z+1}{z^2+1}$

We can see that the function is increasing for , but because of the denominator, it is negative for .

Therefore,

(A) is false because $\dfrac{1}{2} < 1$

(B) is true because

(D) is false, just consider $\ln\frac{z+1}{z^2+1}$ for $z=1$ and $z=2$

7 0

2 years ago

In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean

Andrej [43]

Answer: 1022.75 kWh.

Step-by-step explanation:

Given: In one region, the September energy consumption levels for single-family homes are found to be normally distributed with a mean of 1050 kWh and a standard deviation of 218 kWh.

i.e. $\mu=1050\ kWh$ and $\sigma=218\ kWh$