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4 years ago

What is the greatest prime you must consider to test whether 7066 is prime?

Mathematics

2 answers:

Liono4ka [1.6K]4 years ago

8 0

Answer:

Step-by-step explanation:

The square root of 7066 is 84.0595....

Therefore the largest number we must test is 84, seeing as now we have proven we do not need to test any numbers greater than 84.0595...

That means, we only need to test prime numbers smaller than 84 to see if they go into 7066.

84 isn't prime. But 83 is prime.

Therefore, the greatest prime that needs to be considered for divisibility is 83.

zepelin [54]4 years ago

5 0

Answer:

The numbers 7066 is not prime .

But we easily know when a number is prime when we divide it by the first prime numbers; 2, 3, 5, 7, 11.

Step-by-step explanation:

It is known with certainty that a number is prime when it is only divisible by and by unity.

But in this case, that number, when divided by other numbers less than the one, shows that when dividing into the remainder, it gives zero, which means that it is not only divisible by itself or by the unit, but also by other numbers.

for example; 7066/2=3533 , the remainder is zero;

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Answer:

(3x - 15)/5

Step-by-step explanation:

To find 30% of x+(x-10) we divide x+(x-10) by 100 then multiply it by 30

(x+(x-10)) ÷ 100 × 30 = (3x - 15)/5

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3 years ago

How do you solve the problem 3x+12=15

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3 years ago

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3 years ago

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Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

mote1985 [20]

Answer:

$\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

Step-by-step explanation:

To find the derivative of the function $y(x)=\ln \left(\frac{x}{x^2+1}\right)$ you must:

Step 1. Rewrite the logarithm:

$\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 2. The derivative of a sum is the sum of derivatives:

$\left(\ln{\left(x \right)} - \ln{\left(x^{2} + 1 \right)}\right)^{\prime }}={\left(\left(\ln{\left(x \right)}\right)^{\prime } - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }\right)$

Step 3. The derivative of natural logarithm is $\left(\ln{\left(x \right)}\right)^{\prime }=\frac{1}{x}$

${\left(\ln{\left(x \right)}\right)^{\prime }} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }={\frac{1}{x}} - \left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }$

Step 4. The function $\ln{\left(x^{2} + 1 \right)}$ is the composition $f\left(g\left(x\right)\right)$ of two functions $f\left(u\right)=\ln{\left(u \right)}$ and $u=g\left(x\right)=x^{2} + 1$

Step 5. Apply the chain rule $\left(f\left(g\left(x\right)\right)\right)^{\prime }=\frac{d}{du}\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^{\prime }$

$-{\left(\ln{\left(x^{2} + 1 \right)}\right)^{\prime }} + \frac{1}{x}=- {\frac{d}{du}\left(\ln{\left(u \right)}\right) \frac{d}{dx}\left(x^{2} + 1\right)} + \frac{1}{x}\\\\- {\frac{d}{du}\left(\ln{\left(u \right)}\right)} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- {\frac{1}{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}$

Return to the old variable:

$- \frac{1}{{u}} \frac{d}{dx}\left(x^{2} + 1\right) + \frac{1}{x}=- \frac{\frac{d}{dx}\left(x^{2} + 1\right)}{{\left(x^{2} + 1\right)}} + \frac{1}{x}$

The derivative of a sum is the sum of derivatives:

$- \frac{{\frac{d}{dx}\left(x^{2} + 1\right)}}{x^{2} + 1} + \frac{1}{x}=- \frac{{\left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right)}}{x^{2} + 1} + \frac{1}{x}=\frac{1}{x^{3} + x} \left(x^{2} - x \left(\frac{d}{dx}\left(1\right) + \frac{d}{dx}\left(x^{2}\right)\right) + 1\right)$

Step 6. Apply the power rule $\frac{d}{dx}\left(x^{n}\right)=n\cdot x^{-1+n}$

$\frac{1}{x^{3} + x} \left(x^{2} - x \left({\frac{d}{dx}\left(x^{2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(x^{2} - x \left({\left(2 x^{-1 + 2}\right)} + \frac{d}{dx}\left(1\right)\right) + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x \frac{d}{dx}\left(1\right) + 1\right)\\$

$\frac{1}{x^{3} + x} \left(- x^{2} - x {\frac{d}{dx}\left(1\right)} + 1\right)=\\\\\frac{1}{x^{3} + x} \left(- x^{2} - x {\left(0\right)} + 1\right)=\\\\\frac{1 - x^{2}}{x \left(x^{2} + 1\right)}$

Thus, $\frac{d}{dx}\left(\ln \left(\frac{x}{x^2+1}\right)\right)=\left(\ln{\left(\frac{x}{x^{2} + 1} \right)}\right)^{\prime }=\frac{-x^2+1}{x\left(x^2+1\right)}$

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3 years ago