Ask question

Ask question

All categories

3 years ago

11

Given a 10 foot length of tree trunk with a radius of 3 feet, what is the weight of the tree trunk section if it has a density o

f 45 lb/ft3?

1 answer:

Nutka1998 [239]3 years ago

6 0

treat the tree trunk as a cylinder

v= pi x r^2 x h

using 3.14 for pi

3.14 x 3^2 x 10 = 282.6 cubic feet

282.6 * 45 = 12, 717 pounds

You might be interested in

Help!!!!! Please!!! Limited time!!!! Please thanks

zmey [24]

Answer:C

Step-by-step explanation:

the square root of 12 is 3.46

6 0

3 years ago

Read 2 more answers

In Bonaparte Valley School District, 13% of all students play football and baseball and 25% of all students play baseball. What

vladimir1956 [14]

Answer:

13/25 or 52%

Step-by-step explanation:

7 0

3 years ago

Where does f(x) = 3x2 – 11x – 4 intersect the x-axis? The negative x-intercept is at ( , 0). The positive x-intercept is at ( ,

frez [133]

X=-1/3 and x=4 are the positive and negative x intercepts

6 0

3 years ago

Read 2 more answers

Solve: (-5) + (-18) <br><br>A -23<br><br>B- 13<br><br>C 13<br><br>D 23

Lina20 [59]

The answer would be -23

8 0

2 years ago

In Exercises 5-7, find all the exact t-values for which the given statement is true,

bearhunter [10]

Answer: See Below

<u>Step-by-step explanation:</u>

NOTE: You need the Unit Circle to answer these (attached)

5) cos (t) = 1

Where on the Unit Circle does cos = 1?

Answer: at 0π (0°) and all rotations of 2π (360°)

In radians: t = 0π + 2πn

In degrees: t = 0° + 360n

******************************************************************************

$6)\quad sin (t) = \dfrac{1}{2}$

Where on the Unit Circle does $sin = \dfrac{1}{2}$

<em>Hint: sin is only positive in Quadrants I and II</em>

$\text{Answer: at}\ \dfrac{\pi}{6}\ (30^o)\ \text{and at}\ \dfrac{5\pi}{6}\ (150^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{\pi}{6} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{6} + 2\pi n$

In degrees: t = 30° + 360n and 150° + 360n

******************************************************************************

$7)\quad tan (t) = -\sqrt3$

Where on the Unit Circle does $\dfrac{sin}{cos} = \dfrac{-\sqrt3}{1}\ or\ \dfrac{\sqrt3}{-1}\quad \rightarrow \quad (1,-\sqrt3)\ or\ (-1, \sqrt3)$

<em>Hint: sin and cos are only opposite signs in Quadrants II and IV</em>

$\text{Answer: at}\ \dfrac{2\pi}{3}\ (120^o)\ \text{and at}\ \dfrac{5\pi}{3}\ (300^o)\ \text{and all rotations of}\ 2\pi \ (360^o)$

$\text{In radians:}\ t = \dfrac{2\pi}{3} + 2\pi n \quad \text{and}\quad \dfrac{5\pi}{3} + 2\pi n$

In degrees: t = 120° + 360n and 300° + 360n

4 0

3 years ago