What are the coordinates of the midpoint of,<br> PQ for P(-12, 6) and Q(1,-8)?

A 8-foot tall man is standing beside a 32-foot flagpole. The flagpole is

kipiarov [429]

Answer:

The flagpole's shadow is 16.875 feet longer than the man's shadow

Step-by-step explanation:

The total length of the shadow is expressed by taking its actual length by a factor that depends on the position of the sun which is constant for the man too. The expression is as follows;

Height of the shadow=actual height of the flagpole×factor

where;

length of the flagpole's shadow=22.5 feet

actual height of the flagpole=32 feet

factor=f

replacing;

22.5=32×f

32 f=22.5

f=22.5/32

f=0.703125

Using this factor in the expression below;

Length of man's shadow=actual height of man×factor

where;

length of man's shadow=m

actual height of man=8 feet

factor=0.703125

replacing;

length of man's shadow=8×0.703125=5.625 feet

Determine how much longer the flagpole's shadow is as follows;

flagpoles shadow-man's shadow=22.5-5.625=16.875 feet

The flagpole's shadow is 16.875 feet longer than the man's shadow

7 0

3 years ago

Can someone please help me

Misha Larkins [42]

Try 9cm but I’m not quite sure please reply if it’s wrong or right

6 0

3 years ago

Read 2 more answers

Mathhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

mr Goodwill [35]

It's the first choice.Correlation that implies causation

3 0

3 years ago

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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

3/5x + 3 please help me out

Monica [59]

I’m sorry but there isn’t enough information here to answer anything. What does the equation equal?

5 0

3 years ago

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What are the coordinates of the midpoint of, PQ for P(-12, 6) and Q(1,-8)?