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2 years ago

HELP PLS

Mathematics

1 answer:

slamgirl [31]2 years ago

4 0

The time taken by the bug will be 5 seconds and the formula used to calculate time is T = D/S.

<h3>What is speed?</h3>

The distance covered by the particle or the body in an hour is called speed. It is a scalar quantity. It is the ratio of distance to time.

We know that the speed formula

S = D/T or T = D/S

where T is time, D is distance, and S is speed.

If a bug travels 5 meters across the floor if it is traveling 1 meter/second. Then the time taken by the bug will be

T = 5/1

T = 5 seconds

More about the speed link is given below.

brainly.com/question/7359669

#SPJ1

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What is the area of this circle ( express your answer by using pi)

Leona [35]

Answer: option C

Step-by-step explanation:

You need to use the formula for calculate the area of a circle:

$A_{(circle)}=\pi r^2$

Where the radius of the circle is "r".

You can see in the figure that the diameter of the circle is 6 meters. Then you need to find the radius of the circle with the formula:

$r=\frac{d}{2}$

Where "d" is the diameter of the circle.

Then, the radius is:

$r=\frac{6m}{2}=3m$

Finally, substitute the radius into the formula. Then, the area of the circle is:

$A_{circle}=\pi (3m)^2=9\pi m^2$

7 0

4 years ago

Read 2 more answers

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

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