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3 years ago

The outer layer of the Earth is called the

Mathematics

1 answer:

aksik [14]3 years ago

5 0

Crust is the answer

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Meteorologists recorded a temperature change from 72˚F to 28˚F in one day. Write an equation that represents how many degrees th

marysya [2.9K]

Answer:

72-x=28 or 72-28=x

Step-by-step explanation:

5 0

3 years ago

Can some find the the answer than explain it Thank you

STALIN [3.7K]

Answer:

2 week ago so you dont need so free point

Step-by-step explanation:

7 0

3 years ago

Person 1 writes internal operating procedures 3 times faster than Person 2. Most of the procedures took 30 minutes each to creat

ad-work [718]

Person 1 takes 3 hour 10 minutes to complete 9 procedures

Solution:

From given question,

Person 2 takes 30 minutes per procedure

Person 1 writes internal operating procedures 3 times faster than Person 2

So we get, Person 1 takes 3 times faster than Person 2

Person 1 takes 10 minutes per procedure

But two of the procedures for Person 1 took an hour each

So person 1 takes 60 minutes each for two of procedures ( since 1 hour = 60 minutes )

Calculate how long it took Person 1 to complete 9 procedures:

So for first 7 procedures person 1 would take 10 minutes per procedure and for last two procedures person 1 would take 60 minutes per procedure

$time $=7$ procedures $\times \frac{10 \text { minutes }}{1 \text { procedure }}+2$ procedure $\times \frac{60 \text { minutes }}{1 \text { procedure }}$$

$time=70 minutes +120minutes\\\\time=190minutes$

We know that,

1 hour = 60 minutes

Therefore,

190 minutes = 60 minutes + 60 minutes + 60 minutes + 10 minutes

190 minutes = 1 hour + 1 hour + 1 hour + 10 minutes = 3 hour 10 minutes

Therefore Person 1 takes 3 hour 10 minutes to complete 9 procedures

7 0

3 years ago

PLEEEEEAAAAAASEEEEE HELLLPPPPPP!!! 20 POINTS!!!! list question number then answer

KiRa [710]

A is the correct answer I hope that helps :)

7 0

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