Ask question

Ask question

All categories

3 years ago

15

The distance from the center of a center of a compact disc to the edge of the disc is 6 centimeters. What is the area of the com

pact disc?
A. 452.16 square centimeters
B. 36 square centimeters
C. 113.04 square centimeters
D. 18.84 square centimeters

1 answer:

melamori03 [73]3 years ago

3 0

This is the area of a circle with radius 6 cms
= pi * r^2
= 36pi
= 113.04 cm^2

C

You might be interested in

This data set represents the number of cups of coffee sold in a café between 8 a.m. and 10 a.m. every day for 14 days.

san4es73 [151]

Put the numbers in order...
4,5,6,6,7,8,9,9,10,10,12,12,14,15

Q1 = (6 + 6) / 2 = 12/2 = 6
Q2 (median) = (9 + 9)/2 = 18/2 = 9
Q3 = (12 + 12) / 2 = 24/2 = 12

Q3 - Q1 = 12 - 6 = 6 (this is actually the interquartile range (IQR)

8 0

3 years ago

Read 2 more answers

Which relation is also a function (i can’t get this wrong please help) 15 points

Nuetrik [128]

Answer:

The answer is d

Step-by-step explanation:

this is because a , b and c all have multiple x vales with more then one y value .

6 0

3 years ago

Consider the following differential equation to be solved by undetermined coefficients. y(4) − 2y''' + y'' = ex + 1 Write the gi

kompoz [17]

Answer:

The general solution is

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2}$

Step-by-step explanation:

Step :1:-

Given differential equation y(4) − 2y''' + y'' = e^x + 1

The differential operator form of the given differential equation

$(D^4 -2D^3+D^2)y = e^x+1$

comparing f(D)y = e^ x+1

The auxiliary equation (A.E) f(m) = 0

$m^4 -2m^3+m^2 = 0$

$m^2(m^2 -2m+1) = 0$

$(m^2 -2m+1) this is the expansion of (a-b)^2$

$m^2 =0 and (m-1)^2 =0$

The roots are m=0,0 and m =1,1

complementary function is $y_{c} = (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x$

<u>Step 2</u>:-

The particular equation is $\frac{1}{f(D)} Q$

P.I = $\frac{1}{D^2(D-1)^2} e^x+1$

P.I = $\frac{1}{D^2(D-1)^2} e^x+\frac{1}{D^2(D-1)^2}e^{0x}$

P.I = $I_{1} +I_{2}$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x + \frac{1}{D^{2} } e^{0x}$

$\frac{1}{D} means integration$

$\frac{1}{D^2} (\frac{x^2}{2!} )e^x = \frac{1}{2D} \int\limits {x^2e^x} \, dx$

applying in integration u v formula

$\int\limits {uv} \, dx = u\int\limits {v} \, dx - \int\limits ({u^{l}\int\limits{v} \, dx } )\, dx$

$I_{1}$ = $\frac{1}{D^2(D-1)^2} e^x$

$\frac{1}{2D} (e^x(x^2)-e^x(2x)+e^x(2))$

$\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

$I_{2}$ $= \frac{1}{D^2(D-1)^2}e^{0x}$

$\frac{1}{D} \int\limits {1} \, dx= \frac{1}{D} x$

again integration $\frac{1}{D} x = \frac{x^2}{2!}$

The general solution is $y = y_{C} +y_{P}$

$y= (C_{1}+C_{1}x) e^0x+(C_{3}+C_{4}x) e^x +\frac{1}{2} (e^x(x^2-2x+2)-e^x(2(x-1)+e^x(2))$

+ $\frac{x^2}{2!}$

3 0

3 years ago

Hi, another math addition fraction question, please help.

Irina18 [472]

Answer:

11/15

Step-by-step explanation

7 0

3 years ago

Read 2 more answers

F(x) = 2x^2+ 5x + 20<br><br> Find f(-9)

8_murik_8 [283]

Answer:

137

Step-by-step explanation:

2(-9)^2+5(-9)+20

2(81)-45+20

162-45+20

117+20

137

8 0

3 years ago