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4 years ago

11

WHY is the work they give us online harder than the work we get at school? SOMEBODY PLEASE HELP :(

1 answer:

Dmitry_Shevchenko [17]4 years ago

8 0

Answer:

First question

Circle A:

Radius:2.5 Diameter: 5 Circumference:15.71

Circle B:

Radius:12 Diameter:24 Circumference:75.4

Circle C:

Radius:32 Diameter:64 Circumference:201.06

Circle D:

Radius:42 Diameter:84 Circumference:263.89

Second one:

without Pi

1:81.68

2:37.7

3:263.89

with pi

1:26π

2:12π

3:84π

sorry, i have to keep looking back at the photo

for the last question i will be solving ok

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273.261 in word form

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Two hundred and seventy three point two hundred and sixty one

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3 years ago

Please really need help on this thanks

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3 years ago

Read 2 more answers

Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t> 0 Given yı(t) = t y2(t) = Preview

Anestetic [448]

Let $y_2(t)=tv(t)$ . Then

${y_2}'=tv'+v$

${y_2}''=tv''+2v'$

and substituting these into the ODE gives

$t^2(tv''+2v')+3t(tv'+v)-3tv=0$

$t^3v''+5t^2v'=0$

$tv''+5v'=0$

Let $u(t)=v'(t)$ , so that $u'(t)=v''(t)$ . Then the ODE is linear in $u$ , with

$tu'+5u=0$

Multiply both sides by $t^4$ , so that the left side can be condensed as the derivative of a product:

$t^5u'+5t^4u=(t^5u)'=0$

Integrating both sides and solving for $u(t)$ gives

$t^5u=C\implies u=Ct^{-5}$

Integrate again to solve for $v(t)$ :

$v=C_1t^{-6}+C_2$

and finally, solve for $y_2(t)$ by multiplying both sides by $t$ :

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4 years ago

10 ÷ (20 ÷ 2² × 5 ÷ 5) × 8 - 2

dangina [55]

Use PEMDAS:

P arenthesis
E xponents
M ultiplication
D ivision
A dd
S ubstract

Paranthesis and Exponents
(20 / 2^2 * 5 / 5) = 5

Multiplication
5*8= 40

Division
10 / 40 = 0.25

Subtraction
0.25 - 2 = −1.75

Therefore,
The answer is -1.75

5 0

3 years ago