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3 years ago

Please please use explanation with answer

Mathematics

1 answer:

KIM [24]3 years ago

5 0

Answer:

$\large\boxed{\theta=-\dfrac{\pi}{4}+k\pi\ or\ \theta=k\pi}\ for\ k\in\mathbb{Z}$

Step-by-step explanation:

$\tan^2\theta+\tan\theta=0\\\\\tan\theta(\tan\theta+1)=0\iff\tan\theta=0\ \vee\ \tan\theta+1=0\\\\\tan\theta=0\Rightarrow \theta=k\pi\ \text{for}\ k\in\mathbb{Z}\\\\\tan\theta+1=0\qquad\text{subtract 1 from both sides}\\\\\tan\theta=-1\Rightarrow\theta=-\dfrac{\pi}{4}+k\pi\ \text{for}\ k\in\mathbb{Z}$

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jeremy has a job packing boxes in a warehouse. Jeremy must pack 20 boxes each day that he works to meet quota. He can pack 4 box

eimsori [14]

Answer:

Yes he has enough boxes to meet quota.

Step-by-step explanation:

4 x 8 = 32 This is over what he is requried to meet quota he is at quota.

7 0

3 years ago

Read 2 more answers

Could someone show me how to start this function in excel.

Mrac [35]

Attached is a screenshot of spreadsheet used to do this problem.

You will see the excel functions used for each column. The average score is highlighted in blue.

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3 years ago

Y''+y'+y=0, y(0)=1, y'(0)=0

mars1129 [50]

Answer:

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form $ay''+by'+cy=0$ .

Given: $y''+y'+y=0$

Let $y=e^{rt}$ be it's solution.

We get,

$\left ( r^2+r+1 \right )e^{rt}=0$

Since $e^{rt}\neq 0$ , $r^2+r+1=0$

{ we know that for equation $ax^2+bx+c=0$ , roots are of form $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ }

We get,

$y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}$

For two complex roots $r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta$ , the general solution is of form $y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )$

i.e $y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

Applying conditions y(0)=1 on $e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$ , $c_1=1$

So, equation becomes $y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

On differentiating with respect to t, we get

$y'=\frac{-1}{2}e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )+e^{\frac{-t}{2}}\left ( \frac{-\sqrt{3}}{2} \sin \left ( \frac{\sqrt{3}t}{2} \right )+c_2\frac{\sqrt{3}}{2}\cos\left ( \frac{\sqrt{3}t}{2} \right )\right )$

Applying condition: y'(0)=0, we get $0=\frac{-1}{2}+\frac{\sqrt{3}}{2}c_2\Rightarrow c_2=\frac{1}{\sqrt{3}}$

Therefore,

$y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )$

3 0

3 years ago

HELP me plzzzz i need to finish this

trapecia [35]

Answer:

725.95

Step-by-step explanation:

If you multiply 129.95*5. You get 649.75, but you have to add 76.20 from the delivery and you get 725.95

Hope this helps =)

4 0

2 years ago

Moselle is riding a roller coaster that is 155 feet high.

DerKrebs [107]

Answer:

Step-by-step explanation:

155-16s²

155- 16x2²

155- 16x4

155-64

5 0

2 years ago