All categories

3 years ago

Find the five consecutive Integer have a product of 524,160

Mathematics

1 answer:

ValentinkaMS [17]3 years ago

6 0

The numbers are:

<h2>Explanation:</h2>

An integer is a number that consists of the zero, the positive natural numbers and the negative integers. So, in this case, we have to find the five consecutive Integer have a product of 524,160.

So, let the first number be $x$ then the other consecutive integers can be found as follows:

$x:First \ number \\ \\ x+1:Second \ number \\ \\ x+2:Third \ number \\ \\ x+3:Fourth \ number \\ \\ x+4:Fifth \ number$

So the product is:

$x(x+1)(x+2)(x+3)(x+4)=524160$

We can solve this problem saying that:

$f(x)=x(x+1)(x+2)(x+3)(x+4) \\ \\ \\ Or: \\ \\ y=x(x+1)(x+2)(x+3)(x+4) \\ \\ \\ If \ we \ set \ y= 524160 \\ \\ \\ Then: \\ \\ x(x+1)(x+2)(x+3)(x+4)=524160$

By using graphing tool, we plot:

<u>1. The function:</u>

$f(x)=x(x+1)(x+2)(x+3)(x+4)$

$y=524160$

The solution is the intersection of both the function f(x) and the line, which is x. The result is shown below. So:

$x=12 \\ \\ x+1=13 \\ \\ x+2=14 \\ \\ x+3=15 \\ \\ x+4=16$

<h2>Learn more:</h2>

Cross section: brainly.com/question/12540762

#LearnWithBrainly

You might be interested in

Question is in the attachment

ruslelena [56]

Sorry I can’t see the pic

6 0

3 years ago

Divide by long division method 4x³-x+1by(2x-1)

katen-ka-za [31]

\left[x _{2}\right] = \left[ \frac{-1+i \,\sqrt{3}+2\,by+\left( -2\,i \right) \,\sqrt{3}\,by}{2^{\frac{2}{3}}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{24}+\left( \frac{-1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( 432\,by+\sqrt{\left( -6912+41472\,by+103680\,by^{2}+55296\,by^{3}\right) }\right) }}{\sqrt[3]{2}}\right][x2]=⎣⎢⎢⎢⎢⎡2323√(432by+√(−6912+41472by+103680by2+55296by3))−1+i√3+2by+(−2i)√3by+3√224−3√(432by+√(−6912+41472by+103680by2+55296by3))+(24−1i)√33√(432by+√(−6912+41472by+103680by2+55296by3))⎦⎥⎥⎥⎥⎤
totally answer.

4 0

3 years ago

The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$