All categories

3 years ago

4) A piece of wood in the shape of a rectangular prism has dimensions of 11 cm by 5 cm by 3 cm.

Mathematics

1 answer:

frutty [35]3 years ago

3 0

Answer:

density = 0.58 g/cm^3

Step-by-step explanation:

Mass = 96.4 gram

Volume of rectangular prism = L × B × H

( Where, L = length, B = width, H = height )

=》11 × 5 × 3

=》165 cubic cm

volume = 165 cubic cm

Now, we know that density =

$\frac{mass}{volume}$

=》

$\frac{96.4}{165}$

=》

$0.58 \: g \: cm {}^{ - 3}$

density = 0.58 grams per centimeter cube

( g / cm^3)

You might be interested in

Given:

lisov135 [29]

Answer:

Step-by-step explanation:

6 0

3 years ago

7968 divided by 22 estimating

Anvisha [2.4K]

Answer:

363

Step-by-step explanation:

7 0

3 years ago

What is the discriminant of the quadratic equation 0 = –x2 + 4x – 2? –4 8 12 24

monitta

Answer:

Step-by-step explanation:

0 = –x^2 + 4x – 2

This is of the form

ax^2 +bx +c

a = -1 b = 4 x = -2

The discriminant is

b^2 -4ac

4^2 - 4(-1)(-2)

16 - 8

6 0

3 years ago

Read 2 more answers

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

7 0

3 years ago

Joe wrote the number 2 1/2and 3 1/7 and 1 and 5/6 in the what is a greatest numerador that Joe wrote

Tanzania [10]

3 1/7 because the number is greater.

7 0

4 years ago

Read 2 more answers