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3 years ago

92.3-(3.2 divided by 0.4) times 8

Mathematics

1 answer:

sp2606 [1]3 years ago

3 0

Answer:

28.3 that's the answer

Step-by-step explanation:

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A manufacturer makes ball bearings out of motel steel. it takes 5.24 cubic centimeters of molten steel to make ball bearings tha

katrin2010 [14]

Using volume of ball bearings, a manufacturer makes 10 ball bearings from 5.2 cm³ of molten steel to make ball bearings that each have a diameter of 1 centimeter.

According to the question,

It takes 5.24 cubic centimeters of molten steel to make ball bearings that each have a diameter of 1 centimeter.

Radius = 1/2 = 0.5 cm

volume V of one ball bearing is: V=(4/3)*π*r³

= 4/3 * π * (0.5)³

=0.524 cm³

In order to find number of ball bearings out of molten steel only if we divide the volume of steel by the volume of the ball bearing

5.24/0.524=10.

Hence, using volume of ball bearings, a manufacturer makes 10 ball bearings from 5.2 cm³ of molten steel to make ball bearings that each have a diameter of 1 centimeter.

Learn more about volume here

brainly.com/question/23270880

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4 0

2 years ago

Please help, due tonight

ryzh [129]

Answer:

There's an app call ( m a t h w a y) that would give u the answer in sec

8 0

3 years ago

What is the perimeter of a rectangle with a length of 3.5 yd and a width of 9 yd?

SCORPION-xisa [38]

Perimeter means all the way around.....
so:
3.5 + 9 + 3.5 + 9
= 25

5 0

3 years ago

A suit is on sale for $154. If the original price of the suit was $440, what percent was discounted for the sale? A. 2.5% B. 53%

DochEvi [55]

440 x .65 = 286

440 - 286 = 154

D. 65%

8 0

4 years ago

The indicated function y1(x) is a solution of the given differential equation. use reduction of order or formula (5) in section

Ber [7]

Given that $y_1=e^{2x/3}$ , we can use reduction of order to find a solution $y_2=v(x)y_1=ve^{2x/3}$ .

$\implies {y_2}'=\dfrac23ve^{2x/3}+v'e^{2x/3}=\left(\dfrac23v+v'\right)e^{2x/3}$
$\implies{y_2}''=\dfrac23\left(\dfrac23v+v'\right)e^{2x/3}+v''e^{2x/3}=\left(\dfrac49v+v'+v''\right)e^{2x/3}$

$\implies9y''-12y'+4y=0$
$\implies 9\left(\dfrac49v+v'+v''\right)e^{2x/3}-12\left(\dfrac23v+v'\right)e^{2x/3}+4ve^{2x/3}=0$
$\implies9v''-3v'=0$

Let $u=v'$ , so that

$9u'-3u=0\implies 3u'-u=0\implies u'-\dfrac13u=0$
$e^{-x/3}u'-\dfrac13e^{-x/3}u=0$
$\left(e^{-x/3}u\right)'=0$
$e^{-x/3}u=C_1$
$u=C_1e^{x/3}$

$\implies v'=C_1e^{x/3}$
$\implies v=3C_1e^{x/3}+C_2$

$\implies y_2=\left(3C_1e^{x/3}+C_2\right)e^{2x/3}$
$\implies y_2=3C_1e^x+C_2e^{2x/3}$

Since $y_1$ already accounts for the $e^{2x/3}$ term, we end up with

$y_2=e^x$

as the remaining fundamental solution to the ODE.

7 0

3 years ago