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3 years ago

Dexter ran 5 miles as part of his training. How many yards did he run?

Mathematics

1 answer:

anzhelika [568]3 years ago

4 0

Answer:

8800 yards

Step-by-step explanation:

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You are in a bike race. When you get to the first checkpoint, you are 2525 of the distance to the second checkpoint. When you ge

svetoff [14.1K]

1/10
or maybe even 16?
not too sure though

3 0

2 years ago

Theo can run 1/2 mile in 1/10 hour. If theo continues at this rate how far can he run in a hour?

kirill [66]

1/2 mile x 10 = 5 miles

5 0

3 years ago

Someone help please

vekshin1

Steps to finding the line in the diagram with the format 'ax + by = c

1. Find the slope

To find the slope, we need any two points on the line --> (0,4) and (3,0)

$Slope = \frac{y2-y1}{x2-x1} =\frac{4-0}{0-3} =-\frac{4}{3}$

2. Set up, with any one point on the line and the slope, in point-slope form

$(y-y0)=m(x-x0)\\(y-4)=-\frac{4}{3} (x-0)\\y-4 = -\frac{4}{3} x\\\frac{4}{3}x+y = 4$

<u>Answer</u>: $\frac{4}{3}x+y = 4$

Hope that helps!

5 0

1 year ago

Read 2 more answers

Find the slope of (0, 4), (− 5, 2)

vredina [299]

Answer:

⅖ is the slope

Step-by-step explanation:

To find slope:

$\frac{ {y}^{2} - {y}^{1} }{ {x}^{2} - {x}^{1} }$

$\frac{2 -4}{ - 5 - 0}$

$\frac{ - 2}{ - 5}$

$\frac{2}{5}$

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