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3 years ago

Yusuf walked 4 miles per hour for 0.7 hours. How far did Yusuf walk?

Mathematics

1 answer:

DIA [1.3K]3 years ago

4 0

Answer:

Yusuf walked 2.8 miles.

Step-by-step explanation:

r = 4 miles / hour

t = 0.7 hours.

d = r*t

d = 4 * 0.7

d = 2.8 miles

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I’m confused on your question

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3 years ago

4(-2d-3)=12, someone please solve this for me asap

wlad13 [49]

4(-3 + -2d) = 12

(-3 * 4 + -2d * 4) = 12

(-12 + -8d) = 12

-12 + -8d = 12

-12 + 12 + -8d = 12 + 12

-12 + 12 = 0

0 + -8d = 12 + 12

-8d = 12 + 12

12 + 12 = 24

-8d = 24

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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

You want to place a towel bar that is 24 1/4 centimeters long in the center of a door that is 70 1/3 centimeters wide. How far s

Aleksandr [31]

Answer:

The towel bar should be placed at a distance of $23\frac{1}{24}\ cm$ from each edge of the door.

Step-by-step explanation:

Given:

Length of the towel bar = $24\frac14\ cm$

Now given length is in mixed fraction we will convert in fraction.

To Convert mixed fraction into fraction Multiply the whole number part by the fraction's denominator, then Add that to the numerator, then write the result on top of the denominator.

$24\frac14\ cm$ can be Rewritten as $\frac{97}{4}\ cm$