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3 years ago

What two whole numbers is 29−−√ between?: 3 and 4 4 and 5 5 and 6 6 and 7

Mathematics

1 answer:

jolli1 [7]3 years ago

8 0

The two whole numbers are 5 and 6 ⇒ 3rd answer

Step-by-step explanation:

To prove that a square root number lies between which two consecutive integers do that

Find a square number less than the number under the root
Find a square number greater than the number under the root
Find the square root of the square numbers, they will be the two integers that the root lies between them

∵ The number is $\sqrt{29}$

- Find a square number less than 29

∵ 25 is a square number

∵ 25 is less than 29

- Find a square number greater than 29

∵ 36 is a square number

∵ 36 is greater than 29

∴ 25 < 29 < 36

- Take √ for each number

∴ $\sqrt{25}$ < $\sqrt{29}$ < $\sqrt{36}$

∵ $\sqrt{25}$ = 5

∵ $\sqrt{36}$ = 6

∴ 5 < $\sqrt{29}$ < 6

The two whole numbers are 5 and 6

Learn more:

You can learn more about the numbers in brainly.com/question/9621364

#LearnwithBrainly

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Answer:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=(\frac{2}{7}\overrightarrow{\rm i})*(\frac{1}{r+2s+3t})+(\frac{6}{7}\overrightarrow{\rm j})*(\frac{2}{r+2s+3t})+(\frac{3}{7}\overrightarrow{\rm k})*(\frac{3}{r+2s+3t})$

Step-by-step explanation:

First, let’s check to see if the direction vector is a unit vector:

$||v||=\sqrt{(14)^{2} +(42)^{2} +(21)^{2} } =\sqrt{2401} =49$

It’s not a unit vector. therefore let's divide the vector by its magnitude in order to convert it into a unit vector:

$\overrightarrow{\rm v}=(\frac{14}{49}\overrightarrow{\rm i})+(\frac{42}{49}\overrightarrow{\rm j})+(\frac{21}{49}\overrightarrow{\rm k})= (\frac{2}{7}\overrightarrow{\rm i})+(\frac{6}{7}\overrightarrow{\rm j})+(\frac{3}{7}\overrightarrow{\rm k})$

Now, the directional derivative is given by:

$D_\overrightarrow{\rm v}$ $h(r,s,t)=\frac{\partial h(r,s,t)}{\partial r}\overrightarrow{\rm i} + \frac{\partial h(r,s,t)}{\partial s}\overrightarrow{\rm j} + \frac{\partial h(r,s,t)}{\partial t}\overrightarrow{\rm k}$

So let's calculate the partial derivates:

$\frac{\partial }{\partial r} ln(3r+6s+9t)=\frac{3}{3r+6s+9t}=\frac{1}{r+2s+3t}$

$\frac{\partial }{\partial s} ln(3r+6s+9t)=\frac{6}{3r+6s+9t}=\frac{2}{r+2s+3t}$

$\frac{\partial }{\partial t} ln(3r+6s+9t)=\frac{9}{3r+6s+9t}=\frac{3}{r+2s+3t}$

Therefore:

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