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3 years ago

A cylindrical metal pipe has a diameter of 8.4 millimeters and

2 answers:

vladimir1956 [14]3 years ago

7 0

Volume of a cylinder = pi x radius^2 x height of cylinder and diameter/2 is the radius so...

3.14 x (8.4/2) ^ 2 x 10 = 553.9

Since there is a hole that means you have to find the volume of the hole and subtract that from the volume of the whole pipe (553.9)

3.14 x (6/2) ^2 x 10 = 282.6

553.9 - 282.6 = 271.3

Artemon [7]3 years ago

4 0

Answer:

271.3mm

Step-by-step explanation:

d=8.4

h=10

r=4.2

A=(3.14(4.2^2)) 10

A=554.176944093

d=6

h=10

r=3

A=(3.14(3^2)) 10

A=282.743338823

Now you're going to want to subtract the second area from the first area to get the leftover amount.

554.176944093 - 282.743338823 =271.433661177

~271.3mm

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Please I need help asap

jasenka [17]

Answer:

Option D - x = 3 3/5

Step-by-step explanation:

Lets make the left side into an improper fraction :

2 2/5 = 12/5

Now let's make both fraction into equivalent ones but with the same denominator :

2/3 = 10/15

12/5 = 36/15

Now we substitute the equivalent values back into the equation :

10/15 × x = 36/15

Now we multiply both sides by 15 to remove the denominators:

10 × x = 36

Now we divide both sides by 10 and simplify :

x = 36/10

x = 18/5

x = 3 3/5

So our answer will be Option D

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2 years ago

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3 years ago

Use l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

steposvetlana [31]

Answer:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}$

Step-by-step explanation:

The limit is:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}$

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

$\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}$

by replacing, and applying repeatedly you obtain:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}$