Ask question

Ask question

All categories

3 years ago

10

PLS HELP :(

2 answers:

PtichkaEL [24]3 years ago

6 0

The answer is going to be 47.1

faust18 [17]3 years ago

4 0

Answer: 47.1

Step-by-step explanation:

1/3πr^2h

=1/3×π×32×5

=15π

= 47.123889803847 feet3

You might be interested in

How do you solve this

mars1129 [50]

Answer:

The inquality is always false i think..

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Help pls And thanks

hoa [83]

Answer:

I think that 110 would be the correct answer. PLEASE correct if I'm wrong.

Step-by-step explanation:

See, a 90 degree angle is a right angle or like one of the ways you can split a square in half. l

So since that is a little more than a 90 degree angle, I would guess it would be 110.

6 0

3 years ago

Read 2 more answers

Solve y ' ' + 4 y = 0 , y ( 0 ) = 2 , y ' ( 0 ) = 2 The resulting oscillation will have Amplitude: Period: If your solution is A

Vlad [161]

Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

Now, substitute $y(x)=e^{\lambda x}$ into the differential equation:

$\frac{d^2}{dx^2} (e^{\lambda x} ) +4e^{\lambda x} =0$

Using the characteristic equation:

$\lambda ^2 e^{\lambda x} + 4e^{\lambda x} =0$

Factor out $e^{\lambda x}$

$e^{\lambda x}(\lambda ^2 +4) =0$

Where:

$e^{\lambda x} \neq 0\\\\for\hspace{3}any\hspace{3}\lambda$

Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

$y(x)=c_1 (cos(2x)+isin(2x))+c_2(cos(2x)-isin(2x))\\\\Regroup\\\\y(x)=(c_1+c_2)cos(2x) +i(c_1-c_2)sin(2x)\\$

Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

$y'(x)=2c_1 cos(2x)-2c_2sin(2x)$

Evaluating $y(0)=2$ :

$y(0)=2=c_1sin(0)+c_2cos(0)\\\\2=c_2$

Evaluating $y'(0)=2$ :

$y'(0)=2=2c_1cos(0)-2c_2sin(0)\\\\2=2c_1\\\\c_1=1$

Finally, the solution is given by:

$y(x)=sin(2x)+2cos(2x)$

5 0

4 years ago

Can someone give me some 4th grade math problems

stich3 [128]

1. What is 32×25?

2. What is 856×4?

3. What is 567×40?

Show your work. Sorry if these are too hard.

8 0

4 years ago

Read 2 more answers

Two cyclists , A and B , are going on a bike ride and are meeting at an orchard . They left home at the same time . Functions A

vova2212 [387]

Answer:

Cyclist A lives farther from the Orchard

Step-by-step explanation:

Given

$A(x) =48.5 -21x$

$B(x) =42 -16.8x$

Required:

Who lives farther from the Orchard.

To do this, we simply calculate the y intercept.

This represents the total distance traveled to get to the Orchard

To calculate y intercept, we set $x = 0$

So:

$A(x) =48.5 -21x$

$A(0) = 48.5-21*0 =48.5-0 =48.5$

$B(x) =42 -16.8x$

$B(0) = 42 - 16.8*0 =42-0 =42$

From the above computation:

$A(0) > B(0) \to 48.5 > 42$

Hence;

Cyclist A lives farther from the Orchard

5 0

3 years ago